WO2007014522A1 - A program logic structure graphical representation method and a program path statistic method - Google Patents

Abstract

The program logic structure graphical representation method includes the steps: A. generating a top layer object; B. nesting the branch tree object /statement block object into the branch object of arbitrary layer; C. calculating the size and position of objects in the whole program to draw the graphical representation. The program path statistics method is that the said step C is replaced with program path statistics. The invention may use structural object as plot unit and the size and position of object is automatically calculated, so it is possible to rapidly draw program logic structure graph. The logic structure graph can intuitively represent execution path of program. The program logic structure graphical representation method provided by the invention can automatically perform program path statistics and can design path coverage-based test case according to the statistical result.

Description

 »Logo 3⁄43⁄4 Graphic Method and Road Fiber Travel Brigade

 Specifically, the software development team is involved in the detailed design and re-examination of the field of software development, and more specifically relates to the detailed structure and the logical structure diagram method in the field of testing, and the method of each method in the field of retesting. . Background

 Two important stages of detailed design and software development. Detailed design for the target system «line ratio g fine description; program test is generally divided into black: white box test, which white box test ® ^ internal logic structure design test i test. Although a lot of software is currently oriented (like the whole, but its local code is still structured, detailed design and white-box testing often need to draw a graphical representation of the local logic.

 At present, the commonly used program logic structure diagram method is to draw a program logic structure diagram by using drawing tool software and using the graphic symbols of the logic structure diagram. There are three commonly used logical structure diagrams, » flowcharts, box diagrams (N-S diagrams), PAD diagrams, and their commonly used partial diagrams will be as shown in Figure 2, where Figure 2A is part of the graphical symbol of the f ^ flowchart 201 is a selection (branch), 202 is a process, 203 is a start or stop, 204 is a predefined process, 205 is an input and output, 206 is a control flow; FIG. 2B is a partial graphical symbol of the box diagram, and 211 is a sequence. 212 is a condition, 213 is a loop, 214 is a call subroutine A; Fig. 2C is a partial graph 1 symbol of the PAD graph, 221 is a sequence, 222 is a loop (WHILB C DO P), and 223 is a select 3⁄4^. 111^? 1 518 £ 2), 224 is the definition.

 Figure 3 is a flow chart of the sample code of Figure 1A. »Ber's mapping is a long-standing, arbitrarily described »process tool that has disadvantages that are not conducive to the gradual unfavorable shield structure design, and the structure is not easy to structure.

 Figure 4 is a box diagram of the sample code of Figure 1A. Box graphs have clear functional domains, can be arbitrarily transferred, easy to determine local and ^ [5 Sa's scope, easy to represent nested relationships and levy hierarchy, but when the structure is shorter, the box graph is more difficult.

 Figure 5 shows the PAD diagram of the sample code in Figure 1A. PAD is the English abbreviation of the problem analysis diagram. It uses the control structure of the two-dimensional tree structure diagram. It is easier to translate this picture into a code, and the structure is clear, easy to read and understand. Remember, support top-down, step-by-step design approach. - 3⁄4's logic structure diagram 3⁄4{©3⁄4图3⁄4 ^ draw the logic ^ I diagram, low efficiency, especially when » logic «« money, it takes a long drawing time, modify the logical structure diagram Also annoying than cranes. On the other hand, the obtained logical structure diagram is flawed in the test, it is difficult to describe the » path, «automatically itfM^ path. In the current test work, when the path is to be divided, the control flow graph is generally drawn, which increases the workload. In addition, the control flow graph is not enough to describe the path, and it is not convenient to automatically count the path.

 The white box test is a detailed example of the internal logic design of the code. The side logic M is used to measure the amount of i3⁄4®, and the logic cover is covered with a statement, a decision cover, a cover, and a decision path. Different logical MMs have different test integrity, and path coverage is the most classic problem in white box testing, with good measurement integrity. Since the » contains a loop structure, a very simple one can generate a large number of paths. The complete path is not practical in the actual test work. Therefore, the actual number of loops is generally not considered when counting the paths. In the case of the shield ring body and the non-figure shield ring body, this simplified path is called the z path β. Since the complete path is not practical, the path am we mean is a difficult path.

 Path coverage has a good measurement, but it is rarely in the actual test. The main reason is that the cost of the test is too high. The reason for the high cost of the game is two main points: the lack of automatic statistical path methods and tools; {The branching river can multiply the number of paths, and the number of paths is huge when the program is redundant.

On the road filling, "Software Measurement" (39-40 pages, edited by Shanghai Ai Microsoft Co., Ltd., published by Tsinghua University Press, first edition in April 2004), a statistical method: using the path tree 3⁄43⁄43⁄4 ^ All paths, then start from the root node, one calendar, then back to the root node, the leaf node names that are experienced are arranged in a line, and the path is found. If you try to traverse all the leaf nodes, you get all path of. The book does not detail the specific implementation steps. The path tree is often used to calculate routes in network technology, such as the description!^ The tree-shaped forwarding path of the packet in the network. According to the book; 3⁄4 method, 画 draw the path tree, and try to calculate the path with A, it may be feasible, but the workload is large. We have not retrieved the method of automatically generating the path tree based on the code, and found the tool to automatically generate the path tree based on the game code, and did not retrieve the method of further automatic statistics, nor in the software market. Automatic system A tool for calculating the pay path.

 Regarding the operation of the 3⁄4g3⁄4, the path is very difficult, and the guides are inspected for implementation problems. In the prior art, there is no solution for Yuezi. As mentioned above, the current logic structure diagram method is inefficient, and the resulting logic map has a 3⁄4^ defect when used for »testing. At present, the path test lacks the method of automatic statistical path, and it is also a problem when the number of paths is «1. Summary of the invention

 The technical problem to be solved by the present invention is to address the deficiencies of the above-described program logic structure diagram method, and propose a new program logic structure diagram method.

 Another technical problem to be solved by the present invention is to propose a path automatic statistical method.

 In order to understand the technical problems of the following, the present invention proposes the following technical solutions:

 A » logical structure diagram method, difficult to be in the 'package:

 A. «—a top layer of rubber;

B. nesting branch tree suffixes or statements in any level of 3⁄4x marry: ti _;

 c. Calculate the size of each of ±3⁄4 and ί3⁄4ϊ and draw the illustration.

 It can further include:

 a step for adding a branch to a branch tree;

 a step for deleting a difficult branch in a branch tree sculpture;

 a step for modifying a branch decision;

 The steps for compiling the code of the $t.

 Used to hide the 5? camera port to show the X rubber step; .

 The step used to generate a code from a logic map.

 In order to understand another technical problem, the present invention proposes the following technical solutions:

 A method of unifying the path of ii», ί) ΪΕ , , , 用 代替 代替 代替 代替 代替 代替 代替 代替 代替 代替 代替 代替 代替

 The program structure diagram method proposed by the invention uses the structure X sample as a drawing unit, and the structure object can be stacked to construct a logical structure diagram, and the size and position of the sadness are automatically calculated, and the logical structure diagram can be quickly designed, and the logical structure diagram is convenient. Modified, and the code can be generated in both directions. The logical structure diagram can visually represent the execution path of the Sff, and the US is used to count the ^ path. ● The test does not need to draw another icon, which reduces the workload.

 The program path statistical method proposed by the invention automatically counts the path of the path, and the path coverage test case can be designed according to the statistical result. When the program is more expensive, you can hide the part (hes, ignore the hidden image when counting the path, to compress the path §*, improve the usability of the path S test. .

 The following is a detailed description of the specific method of the present invention with reference to the accompanying drawings:

 1 is a schematic diagram of a structure division and a code slice division of an example code and an example code, wherein FIG. 1A is a sample code, FIG. 1B is a schematic diagram of a structure division of the example code, and FIG. 1C is a schematic diagram of a code slice division of the example code.

2 is a partial apostrophe of a logical structure diagram of the background art. Wherein FIG. 2A is a ®f, part of the flowchart of FIG Jian ^{^:} Number; FIG. 2B is a partial graphical symbols of FIG cartridge; FIG. 2C is a partial graphical symbol PAD of FIG.

 Figure 3 is the flow chart of the sample code of Figure 1A.

 Figure 4 3⁄4⁄4 is a box diagram of the example code shown in Figure 1A.

 Figure 5 is a PAD diagram of the example code shown in Figure 1A.

 Figure 6 is a general diagram of one embodiment of the present invention.

 Fig. 7 is a schematic view showing the internal structure of the image and the structure image tree in an embodiment of the present invention, wherein Fig. 7A is a schematic diagram of the internal structure of the image: Fig. 7B is a branch tree; »Structural diagram, Fig. 7C is a schematic diagram of the internal «shaping of the hemisphere, and Fig. 7D is a diagram of the structure 3⁄4]^3⁄4⁄4.

Figure S is a schematic diagram of an illustration of each structure in an embodiment of the present invention, wherein Figure SA is a schematic diagram of a sub-graph, and Figure 8B is a schematic diagram of another diagram, Figure 8C, Figure 8D, FIG. 8E, FIG. 8F, FIG. 8G, and FIG. 8H are diagrams of the branch tree X image. Schematic, diagram SI is a schematic diagram of a diagram of a statement: ^.

 Figure 9 is a diagram showing the process of drawing a logical structure diagram in the embodiment of the present invention, and Figures 9A to 9D show the steps of drawing a logical structure diagram corresponding to the Figure 1B.

 Figure 10 is a schematic illustration of the dimensions and basic dimensions of each of the embodiments of the present invention.

 Figure 11 is a flow chart for calculating the width and height of each of the three images in an example of the present invention, wherein Figure 11A is a schematic diagram for calculating the width and height of the sample, and Figure 11B is for calculating the width and height of the rubber. Flow intent, Figure 11C is a flow diagram for calculating the width and height of a branch tree.

 Figure 12 is a flow chart for calculating the height of each of the «high branching tree X-like inlet lines in one embodiment of the present invention, wherein Figure 12A is a flow chart for calculating the statement: lfe rubber height, and Figure 12B is a flow chart for calculating the height of the branch. Schematic diagram, Fig. 12C is a flow chart for calculating the height of the branch tree oak, and Fig. 12D is a flow chart for calculating the branch line height of the branch line.

 Figure 13 is a schematic view showing an X-shaped illustration in an example of the present invention, wherein Figure 13A is a schematic diagram showing a division, and Figure 13B is a schematic diagram showing another sub-illustration. The branch label is added, FIG. 13C is a schematic diagram showing a branch tree X image, and FIG. 13D is a schematic diagram showing an i-sample illustration.

 Figure 14 is a schematic illustration of a partially illustrated modification of a hidden image in accordance with one embodiment of the present invention.

 Figure 15 is a schematic illustration of a 3⁄4f3⁄4 chip of ^ in another embodiment of the present invention.

 16 is a schematic flow chart of a system path in another embodiment of the present invention, wherein FIG. 16A is a schematic flowchart of a statistical path when the rubber is a block, and FIG. 16B is a flow chart of the statistical path when the object is a branch. 16C is a schematic flow chart of the statistical path when the object is a branch tree. .

 Figure 17 is a schematic diagram showing the iiff path.

Detailed description.

 This male case divides the logic structure of the drunk into two types: sequential structure and structure. The bi structure corresponds to a sequence of 驹. In this sequence, if the first statement is executed, then if there is no exception and no return is returned in advance, the 4th statement will be executed. All consecutive statements of a condition should be considered to belong to the same sequential structure and cannot be split, that is, one part of + cannot be regarded as belonging to one), and the other part is regarded as belonging to another structure. The structure has an entry point and an exit point, the entry point is the first statement in the structure, and the exit point is the last statement in the structure, but a statement in the structure of the chrysanthemum is a return statement (such as

The return language of C++, the function returns when the statement is executed, and the following statement will not be executed. In fact, the following 驹 is invalid, so the statement is the actual exit point of the 3⁄4«* structure. Because GOTO statements (such as C++'s goto breaks down the structural design and are less common in today's development, this example does not consider GOTO statements.

 The branch structure corresponds to the sequence of the sentence. In this "i-my sentence sequence, there are functions of judgment and/or hopping. ^^, under different conditions, some of them may or may not be executed or repeatedly executed ^ The most typical branch structure; 3⁄4 select structure, the loop structure also belongs to the branch structure. Because the branch structure will execute or not execute some statements under different conditions, the branch structure has two possibilities. The execution route, the 3⁄43⁄43⁄4 line is called the 4 branch, that is, a ^3⁄4 structure contains at least two branches. To simplify the problem, the loop structure is considered to have two branches of the loop body and the non-ax loop body, regardless of the loop.

 The more common selection structures are:

 Positive structure, such as C/C "HVJava if(...).Jelse...; Pascal's if... then-else...; Basic IF...THE ...END IF; This structure ELSE branches can be included or not, and an unlimited number of ELSE IF branches can be added before the ELSE is divided into 3⁄41.

 SWITCH structure, such as C/C++/Java switoh(...)..., Pascal case...of..., Basic SELECT CASE..., this structure can contain multiple CASE branches, which can contain It is also possible to exclude the DEFAULT branch.

 The more common loop structures are:

FOR structure: For example, C/C"HVJ _ava for(...)..., Pascal for...do..., Basic Κ)Ι..·ΜΒΧΤ and FOREACH.._ ECT. WHILE structure : such as C/C++/Java while(...)..., Pascal's wMe~do..., Basic's DO WHILE..丄OOP, DO UNTtL..LOOP.

 Do.—WHILE structure: such as C/C++/Java do-while(...), Basic DO....LOOP WHILE..., DCL. LOOP UNTIL.., the structure executes the loop body at least once.

 The above examples of branching structures and programming languages are not exhaustive, and any code structure having two or more execution routes is considered a branching structure.

 ±j , the loop structure is considered to have two branches of 3⁄4 loop body and not loop body, regardless of the loop», and the DO...WHILE loop structure at least one body of the shield ring, therefore, its branch without loop body It is Kodak.

 The branch structure also has an entry point and an exit point, regardless of its internal execution route, in the absence of an exception, the structure is always ax by the entry point, and the exit point leaves the structure, 3⁄4a has ^ If the structure contains a return statement, the function returns when the statement is executed, and does not branch off from the exit point.

The 3⁄4^ branch of 3⁄4^^ is implicit, such as the IF structure. If there is no ELSE branch afterwards, the shell contains an implicit branch, even if it contains an ELSE IF branch. The execution of this implicit branch is The route is: When the IF and all ELSE IF decisions are false, jump directly to the exit point of the structure. The case with implicit branches is also: The loop structure does not enter the loop body branch, the execution conditions and lines of the branch

Structure, when there is no DEFAULT branch, ₇ also contains an implicit branch. The execution route is: all the cakes of the CASE branch are not satisfied, the CASE branch is not executed, and the exit point of ^ I is directly jumped.

 The branch structure consists of at least two branches. It can be said that the branch structure is composed of branches. The branch contains two aspects: branch decision and branch, the calculation result of branch decision generally has only two values: true and false. When the judgment is true, the branch is executed, otherwise the branch is not executed. The branch decisions and branches of the implicit branch are empty. For a certain point, its branch occasionally has a certain range, called scope, and the specific symbol of i3⁄4J3⁄4 defines the beginning and end of the scope, for example: C/C++ with curly braces ({})' PASCAL Begin and end specify the start and end of the scope. All statements in the scope, regardless of their structure, the number, ± hook belongs to the branch, therefore, the branch of a branch can contain one or more 11-order structure, or can contain one or more branch structures, we Called a substructure, the substructure can in turn contain its own substructure.

 In summary, the embodiment divides the logical structure into a branch structure and a sequence structure. The branch structure includes at least two branches. Each branch may further include a structure and/or a branch structure. There are no restrictions on the number of nested levels and numbers. Both the branch structure and the sequence structure have an entry point and an exit point. When the ®m line is to any structure, the structure is always removed from the entry point without leaving an exception, and the structure is removed from the exit point, unless Execute to return and return early.

 This embodiment uses a structure 3⁄4⁄4* (the structure should be described as follows: a branch tree is used to describe the branch structure, and a branch is used to describe a branch in the branch node, using the statement: ^ rubber structure. Among them, the branch tree (The oak nests at least two points 53⁄4, the points 3⁄43⁄4^ can be nested in any number of branch trees 2 / or statement: ^ rubber, the amount of sugar and the number of sugar is not limited, the statement rubber is not embedded. In addition, this embodiment The purpose of the logical structure to be shown is hard to be regarded as a branch, called the top branch, which is described by a top-level rubber. The top layer is 3⁄4 (the oak is different from the other 3⁄4 rubbers, 'it is not embedded*ΐ^ 3⁄4. The latter leg "divided oak", including the other levels of the top tiered guard, 3⁄43⁄4^ we put the branch tree) (he, the oak, the statement: 3⁄43⁄4tm ^ called the structure pulse

 For example, Mm A nested X Yu B, and our Oak A is called the Eagle B. The X-Blane B is called the earner A. The top layer of the rubber is not like the image. The statement: ^There is no child 3 rubber, and the other is 33⁄4. There is no child carving. Like the relationship of the internal faces, (Acorn A (like) nested) Oak B (Zi is indeed, is to withdraw the Oak A to save the Carving B itself. A saves the pointer of the Carving B or the bow I, §3⁄43⁄4#Α Saved the «3⁄4, 3⁄4 3⁄4 like A can be used for 2 rubber 3! With 2 rubber B; Λ3⁄43⁄4 icon, object AC m#) «3⁄4 B (sub is B at least one The direction is within the range of the image A.

 Fig. 7 is a schematic view showing the internal structure of each structure in the present embodiment, wherein Fig. 7A is a schematic diagram of the internal structure of the statement: Fig. 7B is a schematic diagram of the internal surface of the branch tree, Fig. 7C. It is a schematic diagram of the internal turning of the image. In FIGS. 7A to 7C, the solid line shows a fixed component, and the dotted line shows an optional and unlimited number of components. FIG. 7D is a diagram. ^" Schematic, corresponding to the » ^ code as shown in Figure 1A.

 Figure 7Α is a schematic diagram of the internal surface of the statement ±33⁄4 image, as shown in Figure 7, the image has a ~ code field, which is used to store its corresponding » (weight, and has a name field for identifying different statements) ± fine image, the name can be letters, such as ', V,

V. '- Figure 7B is a schematic diagram of the internal structure of the branch tree image. As shown in Figure 7B, the branch tree is nested with at least two sub-S3 images. Take more points and divide. Each branch of 5^3⁄4 is actually a component of the partition of the branch structure. There is no branch, and the branch structure is not. The branch is treated separately as a single, just for the convenience of description and realism. At the same time, it is more in line with the idea of orientation.

 Fig. 7C is a schematic diagram of the internal surface of the image. As shown in Fig. 7C, the sub-word is deleted and the corresponding branch decision is saved. The top branch has no corresponding branch decision, so the field is empty. The scale image can be nested with any number of branch trees Liu / i. If there is no nested branch tree statement: ^ like, then the branch is called an empty sample. Like the oak, the branch has a name field. For different branches 3⁄4m, the names can be represented by letters, such as V, V, V.

 Figure 7D is a schematic diagram of the structure pair corresponding to the code shown in Figure 1. As shown in Figure 7ϋ, all the structures describing the logical structure of a program are turned into a tree, which we call a tree.

 In order to traverse ±3⁄4 each structure pulse 3⁄4h3⁄4 structural oak for post-sequence scanning, the posterior cerebral drawing refers to: Calculate from the top layer 3⁄4 rubber, calculate each rubber for each gland, calculate the sub-carving and calculate The X rubber itself, since the ± image and the space division 3⁄43⁄41^ have no children ^, calculate them directly when calculating themselves, so they are terminated. The first difference is that compared to the post-drawing, the only difference is that for each, the former ^3⁄4 describes itself, and then recursively computes the sub-d.

 The structure Pan has an internal flip and also has a corresponding external display that is displayed on the screen to save his visual media. The combination of the 3⁄4^ icons constitutes the logic of »! Figure.

 Figure 8 is a schematic view of the structure of the present embodiment (the illustration of the rubber. 3⁄4^ illustrated i3⁄4i for the convenience of description and invention, its magic, characters, colors 3⁄43⁄4 his components are not 3⁄4⁄4 to the present invention rights *3⁄4 real » "The definition of the formula, the text and color of these illustrations can be modified as needed when implementing the invention.

The figure is a schematic diagram of a sub-graph, as shown in FIG. 8A, 811 is an execution route of a branch, and a branch tree or a statement block 3⁄4# can be added on this line, and the branch point, the entry point S12, and the exit point S13 are executed. Both ends of the route. The entry point of the top branch is the entry point of Wo, and the exit point of the top branch is the exit point of Wo. - Figure 8B is a schematic diagram of another diagram of the division, and the branch annotation is added on the basis of Figure 8A. As shown in Fig. 麻, 822 is a branch note, and 823 is a branch mark. Branches are displayed in the branch decision or other explanatory text. The width of the branch label is the same as or slightly shorter than the width of the branch. The branch range can be seen from the branch label, so another function of the branch label; In this embodiment, the branch decision is used as the main character of the branch, and the syntax of the language is 3⁄4⁄4C or C++, and different programming languages can be finely corresponding to the syntax. The sub-sum is generally not a worm, but is a component of a branch tree ^, as shown in Figure SC. ₄ Fig. 8C. is a schematic diagram of a branch tree object, as shown in Fig. 8C, S31 is an entry line, and 832 is an exit line. * The branch tree object contains at least two points (掾, which is part of the branch tree sculpture, and the dotted line box 833 and the dotted line box 834 are the two points of the branch tree image «^. 53⁄4. The execution line is: from the entry line 3⁄4, executing a branch, reaching the exit line and leaving the branch tree object. Figure 8D shows the entry and exit points of the branch tree, the entry point 835 and the exit point 836 are respectively located at the entrance The midpoint of the line and the midpoint of the exit line.

 Figure 8A, Figure 8F, Figure 8G Figure 8H ± is a schematic diagram of the branch tree. Since there are many branch structures, different types of branch structures have different numbers of branches, and the branch decisions are different. According to the language used by the editor, the branch that is commonly used in each language - a corresponding branch tree X rubber, for example, can be C++: for, while, if, e, if/else if/else The switch ^^ branch tree ^ is called "for branch tree", "while branch tree", "if branch tree", "if/else branch tree", "if/else if/else branch tree", "switch branch" Tree". The various sub-branched tree oaks differ only in the branch of the component and the branch. The branch tree that should be spoofed in SWITCH has peripheral code. For example, the peripheral code of SWITCH in C++ is switoh(...) {... }. This 3⁄43⁄4 example adds a "f stay" to the branch tree corresponding to the SWITCH structure; the field of the code is called the peripheral code field, and the peripheral code of staying » is used for switeh(...); The class and terminator, the branch tree we should be in the SWITCH structure is called the medical branch tree.

 Figure S [is a diagram of the 3⁄43⁄4 diagram of Figure ,, as shown in Figure 81, 843 is the range rectangle of the statement ± 豫, statement ^ (sample entry point 841 and exit point 842 are located on the left 3⁄4Π3⁄4" direction of the center line and the range rectangle The intersection of the vertical lines at both ends.

 The steps shown in the figure are described in detail below.

 As shown in Figure 6, step A generates a top-level partition, which is the top-level x, and the other steps are directly or indirectly nested in the top-level 3⁄4m.

Figure 6 household; ^, step B at any level of the branch ^ (like nesting branch tree or statement first, you can nest branch tree 3⁄4mn / or statement in the top layer of rubber as needed: ^ inch image, by «- The branch tree) image contains at least two partial images, so Generated other levels of points ^! Oak '3⁄43⁄4 rubber can be 3⁄4^ frequency set branch tree X Royal / or 3⁄4 (Oak.

The present invention can be used to calculate a logical structure diagram (for example, for detailed design), and can also be used to generate a logic map according to an existing » (for example, for a test phase). This male example is used to set up the logical structure diagram, the difficult branch tree in step 8 (the 驹:^ is specified by the user, the branch tree is paired with ^ 3⁄4/ or the statement

And nested on the ^ ^ 3⁄4 of ^ S, you can "± 隹 "" to the logical structure diagram, the number of the number and the level of «3⁄4 is not limited. The following is an example of the process of drawing the »logical structure diagram with m, detailing step B of the cost example:

 Figure 9 is a schematic diagram of the process of drawing a logical structure diagram, and Figures 9A to 9D show the steps of drawing a logical structure diagram corresponding to the code shown in Figure 1B, which are composed of two parts: Above is the external 3⁄43⁄4^ of the logic ^ diagram, the right side is the phase

^ (The internal « like its nesting, the "repainting" of the household M is the step C, which will be described in detail later.

 In order to facilitate the user to select 3⁄4⁄4, the embodiment also provides a 3⁄4*3⁄4 selection interface, which switches the programming language, and displays an icon corresponding to various branch structures (ie, a refined branch tree object) and corresponding to each language. The icon of the jl sequence structure (that is, the statement block is for the user to select. For the convenience of description, when the user selects an icon corresponding to a certain structure, we call to select a certain branch tree, such as a while branch tree, an ififelse branch Tree, when the user selects the icon corresponding to the jl structure, we call it the query block, and the 3⁄4 « is called the selection ^ ^ carving. '- .

 As shown in Figure 9A, the user selects the while branch tree and clicks on the icon of the top-level branch, which corresponds to the code shown in dashed line 101 in Figure 1B. 901 is a schematic diagram of the inner surface of the top layer X, and in the top of the top, the β while branch tree image 902 is redrawn, and the logical structure diagram of the left side of Fig. 9B is obtained.

 As shown in Fig. 9B, the user selects 1:^:^ and clicks on the right side of the top branch object execution line located on the right side of the while branch tree, which should be in the code of the dashed box 105 in Fig. 1B. At the top of the 3⁄43⁄4 internal «中^6 3⁄4^

903 and redraw, get the logical structure diagram of Figure 9C left;

 As shown in Figure 9C, the user selects the if/else branch tree and clicks on the first branch of the while branch tree, which corresponds to the code of the dashed box 102 in Figure 1B. Nesting an «e branch tree 904 in the internal drawing of the first image of the while branch tree and redrawing it, and obtaining the logical structure diagram of Fig. 9D j; jij^.

 As shown in FIG. 9D, the user selects ^±^ (like; and clicks on the first branch and the second branch of the i se branch tree respectively. In the internal β of the second branch of the first branch of the ifelse branch tree, respectively β- My sentence; 3⁄43⁄4] 905, 906 and redraw, get the final 3 series structure diagram. 驹 ± sample 905 corresponds to the code of the dotted frame 103 in Figure 1B, the statement: the rubber 906 corresponds to the dotted line frame 104 in Figure 1B Code. ' - This 3⁄43⁄4 case respectively 価 an auto-incrementing letter »* as the name of i#iOi«^i3⁄4, such as 'a', V, V. Add 驹± ^! when the oak is the 驹 (the name of the oak When adding a branch of Mayan, name it 3⁄4».

 As shown in Fig. 6 , step C calculates the size and health of each 5! and draws the icon. The rear M "repainting" or "repainting logic" refers to the heavy surface Co step C includes three aspects. : Calculate the size of each X image, the cost of the shed and the height of the shed (the size of the image, because the calculation of the height of the branch line of the branch tree is more than »), therefore, the height of the Λ line is also used as the size of the branch tree sculpt in this embodiment. ; Calculate the position of each object, this 3⁄4⁄2 case uses the coordinates of the entry point to 3⁄4 ^ the position of each object; draw a diagram of each object.

 Due to the submarriage, the range is within the range of Yu, therefore, the addition of 3⁄43⁄4ffl, except for one X, will change the size of the direct or inter-sample, and a 3⁄43⁄4 of the health will also change; ^, so, increase I3⁄4W In addition to a X rubber, it is generally necessary to calculate the size of each X rubber and to calculate the relevant size and health of each pan, it is also necessary to predefine a SS size.

 Fig. 10 is a view showing the size, the health and the basic size of each X image in the present embodiment.

 The width and height of the eagle are as good as the width and height of the rectangular shape of the rectangle (called the shape of the rectangle). As shown in FIG. 10, the rectangle 1001 drawn by the dashed line is the range rectangle of the top branch, the rectangle 1002 drawn by the dashed line is the range rectangle of the if/else branch tree, and the rectangle 1003 drawn by the dashed line is the first branch of the while branch tree. The rectangle of the range, the rectangle 1004 drawn by the dashed line is the range rectangle of the first branch of the if/dse branch tree, and the rectangle 1005 drawn by the solid line is an i query ±^5 (the range rectangle of the oak).

 As shown in Figure 10, E1 is the entry line height of the while branch tree object, and E2 is the entry line height of the if/else branch tree object.

 As shown in Figure 10, 1011 is the entry point of the top-level rubber, 1012 is the branch tree (the entrance point of the oak, 1013 is the entry point of the first branch of the while branch tree, and 1014 is the entry point of the oak, their The coordinates are the coordinates of the entry point of the phase i3⁄4 rubber.

 As shown in Figure 10, the beginning of the letter S in Figure 10 is 3⁄43⁄43⁄4 in size, including:

Statement: ^ width of anger S1, 驹: height of ^ image S2; The height of the branch mark S3, the distance between the branch mark and the branch S4, and the distance between the two branches S5; wherein the branch mark is not necessary, if the branch is 3⁄43⁄4, both S3 and S4 can be set to 0, no need to change » 3⁄433⁄4 calculation formula;

 The distance between the two 3⁄4^ of the left 3⁄4^ direction is S6.

 The size varies regardless of the addition and deletion of the sample, but the size of the ship can be scaled.

 The following is a detailed description of step C.

 C1. Calculate the size of each rubber. As before, the dimensions of each carving include: X width and height of each spoon, calculate each) (The steps of width and height of the oak are: starting from the top 3⁄43⁄4, for each carving, recursively calculate each child (働 width) And height, calculate the width and height of the 3⁄43⁄4 width and height.

 Figure 11 is a flow diagram of the width and height of each X image in the present embodiment, wherein Figure 11A is a flow chart for calculating the width of the image: ^ (the width and height of the image, and Figure 11B is the width and height of the calculated score! Schematic, FIG. 11C is a schematic diagram of calculating the width and height of the image of the branch tree 3.

 As shown in Figure HA, t i ., according to the following formula: width and height:

Width = S1 _;

• Height = S2 _;

 Where S1 is the width and S2 is the height of ±^1. '

 For the 3⁄43⁄4, calculate the width and height according to the following formula:

 Width = S1 +2*S6 (N=0);

 Width = ∑W + (N+1)*S6 (N>0);

 Height = S3 + S4 (N=0);

 Height = Hm + S3 + S4 (N>0);

 Where N is the number of sub-objects of the division 3⁄4m, SI is the width of the statement: ^ image, S6 is the distance between the two 3⁄4^ of the left 3⁄4 direction, ∑W is the sum of the widths of all the sub-s, and S3 is the branch note. The height of S4 is the distance between the branch and the branch, and Hm is the height of the sub-carving with the highest height. ,

 Specifically, as shown in Fig. 11B, if the number of sub X images N is 0, the formula is used: width = S1 + 2 * S6, height = S3 + S4 'calculates its own width and height, otherwise the scan is if the sub is Statement: The calculation of the Fig. 11A hemp is performed, otherwise the calculation shown in Fig. 11C is performed recursively; all the sub-biliary calculations are completed, and the sub-carving with the highest height is found, and then according to the formula: Width = ∑W + (N+1) *S6, Height = 13⁄41 + 33 + 34 Calculate its width and height. -.

 For the branch tree, calculate the difficulty width and height according to the following formula:

 Width = Wm;

 Height = ∑H + ( -1) * S5;

 Where Wm is the width of the largest sub-object, ΣΗ is the sum of the heights of all sub-objects, N is the number of sub-objects, and S5 is the distance between the two branches.

 Specifically, the calculation is shown in Fig. 110, the scan is X, and for each sample (sub-X is hesitant), the calculation shown in Fig. 11B is performed recursively, ^^ finds the branch with the largest width, and then calculates according to the formula Its own width and height.

 After calculating the width and height of the branch tree, it is also necessary to make the branches of the branch tree equal in width. In this way, the width of the branch other than the largest branch is removed, so that it is equal to the width of the branch with the largest width.

 As mentioned earlier, the size of each carving can also be entered into. ^Includes: The height of the entrance line of the branch tree carving. The following describes the procedure for calculating the branch line height of the branch tree.

 In order to calculate the entrance line height of the branch tree, it is necessary to calculate the height of each base first. The height of the seat is the distance from the mouth of the 3⁄41 elephant to the bottom edge of the range rectangle, each with its own / height, and the height of β is also the basis for calculating the coordinates of the entry point. After the ff3⁄4 method is used to calculate the height of each test, starting from the top layer, the height of each sub-sample is recursively calculated for each pulse, and the height of the vine is calculated according to the height of each sub-carving.

FIG. 12 is a flow chart for calculating the height of the entrance line of each of the three male and female branches in the present embodiment, wherein FIG. 12A is a flow chart for calculating the height of the 驹±夬X-like height, and FIG. 12B is a flow chart for calculating the height of the vine. 12C is a schematic flow chart for calculating the height of the branch tree, and FIG. 12D is a flow chart for calculating the height of the entry line of the branch tree. - As shown in Figure 12A, ^ i -, the height is calculated according to the following formula: / ^ Height = S2/2 _;

 Where S2 M^ m

 For the image, it is difficult to get away from the height according to the following formula:

 Flight altitude =0 (N=0);

 Vine height = Bm (N>0);

 Where N is the number of sub-X rubbers, Bm is; ^ is the height of the child X image; height.

Specifically calculated as the 1$ port shown in Figure 12B, if the number N of the child is 0, the formula calculates the male height, otherwise the scan, if the child is a statement block, recursively performs the calculation of the graph i2A Bi3⁄4, otherwise recursively Perform the calculation shown in Figure 12C,

After the calculation is completed, the child height is the highest.

 For branch tree carving, calculate its impurity height according to the following formula:

 Hertz height = [Βι + (Ηη-Βη) + Σ(Η2..Ιϊη-ι) + (N-1)*S5] / 2 + Bn;

 Where Bi is the second ^ (the height of the oak is ' π is « the height of the sub X image, Bn is the height of the sub-Yu clock, and Σ (Η 2... Η η-ι) is the other than the "" The sum of the heights, the X is the child X rubber S5 is the distance between the two.

 The specific calculation process is shown in Fig. 12C. Scanning the image, for the ^^3 oak, recursively performs the calculation of Fig. 12B, and all the subsamples are calculated; the shed di formula calculates its own female height.

 As shown in Fig. 12D, the branch tree ^ entry line height is calculated according to the following formula:

 Entrance line height = Βι + (Ηη - Βη) + Σ(Η2...Ηη-ι) + ( - 1)*S5;

 Among them, Bi is the first oak) 3⁄43⁄43⁄4 degrees, and Ηη is «子5! The height of the rubber, Bn is the height of the scorpion, and ∑ (H2..JHn-1) is the sum of the heights of the other 3⁄4m except for the last and last, and N is the distance between the two S5.

 C2. Calculate the β of each X image. As before f3⁄4, each rubber's Zhao includes: the coordinates of each rubber entry point. The steps for calculating the coordinates of each 3⁄4# entry point are as follows: Starting from the top level, acupuncture is sculpted, the coordinates of the entry point of M are calculated first, and the coordinates of the entry points of each sub-port are calculated recursively according to the position of the fiber.

 For ease of description and implementation, the present embodiment uses the following coordinate system: «I coordinate is positive to the right, positive is positive to the bottom, and the origin is the top ± branch of the rectangle of the engraving range. ' .

 For the top-level branch X rubber', calculate the entry point coordinates according to the following formula:

 X=0;

 Y-H- B;

 ■ Where X is the abscissa of the entry point, positive to the right, Y is the entry point 3⁄4 ^ mark, down is positive, H is (health height, B is);

For the N-th image, the coordinates of the dragon entry point are calculated according to the following formula: - Xn=∑(Wi...Wn-i) + N*S6 _;

 Yn=Yi; - where Ν is the sub-oil «^ number from 1 ,, 3⁄41 is the horizontal mark of the entry point of the Ν ΐΧ rubber, positive to the right, ∑ (Wi...Wn-i) is 3⁄4 ^ ^3⁄4 ^ The sum of the widths of the preceding sub-heals, S6 is the distance between the left X⁄4Γ3⁄4 "to the two Xifes, Yn is the 3⁄4^ of the entry point of the household 3⁄4ΜΝ, and the direction is positive, ΥΪ is the minute ^ ^ entry point «mark.

 For the branch of the branching paste, calculate the coordinates of the entry point according to the following formula:

 Xn = Xi;

 Yn=Yl - E/2 (N=l);

 ' where N is the child starting from 1), η is 3⁄43⁄4 N " ^X image of the entry point of the abscissa, right is positive, ΧΪ is the branch tree (like the entry point abscissa, Υη 3⁄4 Ν ^ The mark of the entry point of 3⁄43⁄4 is positive, Yn-i is the front entry point», Bn-i is the β height of the front D rubber, Ηη is the height of the 3⁄43⁄4 Ν ^ image, and Bn is the weave N ^^ Height, S5 is the distance between the two.

 C3. Draw a diagram of each 。. The steps for drawing each 3⁄4 ^ diagram are as follows: Starting from the top layer 3⁄43⁄4, for each X image, first draw an icon of the image, and then recursively draw the icon of each child 3⁄4.

 Before the top layer is divided into 3⁄43⁄4, remove the old whole picture, and draw the picture inside the rectangle of the range of «3⁄4—3.

Each carving has an exit point, indicating C o, Yo): Xo=Xi +W;

 Yo = Yi;

 Where Xi is the abscissa of the entry point, w is the width of the object, and Yi is the ordinate of the entry point.

 Fig. 13 is a schematic view showing each of the drawings (Fig. 13A is a schematic diagram showing a division, and Fig. 13B is a schematic diagram showing another divisional diagram, adding a branch feft Fig. 13C is a schematic diagram showing a branch tree diagram, and Fig. 13D is a diagram showing a ±^3⁄41* diagram.

 As shown in Fig. 13A, the illustrations are divided into: Execution line 811. The method of drawing the execution line 811 is to draw a line segment between the entry point 1311 and the exit point 1312.

 As shown in Fig. 13B, the illustration of another branch image further includes: a branch label 822 and a branch label text 823.

 In addition to drawing the execution line, draw the branch 3⁄43⁄4. In this example, use the rectangle as the branch label. The steps for drawing the branch label are as follows: Calculate the branch giant shape 823, its 3⁄4± angle 1321 coordinates (Χ,Υ) are:

X=Xi + S6 _;

Y=Yi + H -B + S3 + S4 _;

 Where Xi is the abscissa of the entry point of the branch, S6 is the distance between the two directions of the left direction, Yi is the entry point of the branch, H is the height of the branch, B is the height of the branch, and S3 is the branch mark Height, S4 is the distance between the branch and the branch. The width and height of the fractal are:

 Width = W- S6*2;

Height = S3 _;

 Where W is the branch width, S6 is the left; ίΓ ^ is the distance between the two 3⁄4^ connections, and S3 is the height of the branch note.

 In this embodiment, the branch label is not drawn with a border, and only a gray background is drawn.

 Finally, the main text of the branch is output in the branch giant.

 As shown in Fig. 13C, the illustration of the branch tree includes: an entry line 831 and an exit line 832, the drawing is:

 Calculate the coordinate values (XI, Υ1)' (X2, Y2) of the two endpoints of the entry line 831:

 Xl = X2 = Xi; where Xi is the entry point abscissa:

 YI = Υι -Ε/2; where 3⁄4 is the ordinate of the entry point and Ε is the height of the entry line;

 Υ2 = YI +Ε;. where Y1 is the upper endpoint of the entry line S^, and Ε is the entry line height.

 By shifting the entry line to the right by the width of the branch tree?, you can get the exit line 832 with the coordinates:

(X1+W, Y1), (X2+W, Y2) _; where W is the width of the branch tree object.

 Draw a line segment between (X1, Y1), (Χ2, Υ2), which is the entry line 831.

 Draw a line segment between (X1+W, Y1), (X2+W, Y2), which is the exit line 832.

 As shown in Fig. 13D, this embodiment uses the range moment as an illustration of a block. Draw a statement: The method of the 3⁄4X image is:

 Draw a range rectangle with a straight line 843, the rectangle: £± angle 1341 coordinates are:

X = Xi _;

Y = Yi - S2/2 _;

 Where i is the horizontal coordinate of the entry point of the statement block, Yi is the ordinate of the entry point of the statement block, and S2 is the height of the statement block object.

 After the X images are drawn, the obtained graphics are the logical structure diagrams. In order to compile the logical structure diagrams, the following steps can be added in this embodiment:

 The step of adding a branch to the branch tree image, the step specifically includes:

 1) ■ hit test 3⁄4 ^ method to determine which branch tree the user has selected

2) Add "^ points 3⁄43⁄4" to the branch tree _;

 3) Redraw the logical structure diagram.

 The hit test is the method of judging whether the user has selected a 3⁄43⁄4⁄4 mouse to stay on a certain location from the position of the mouse or the cursor. In this embodiment, the hit test method is used to perform the hit test: starting from the top-level scoring, each sub-recursive test is performed for each recursion, and the column test is performed, and the test is ended when a certain engraving is hit, and the method for judging the hit is: The mouse or cursor position is within the bounding rectangle of the gallbladder and hits, otherwise ». '·

The step of deleting the selected branch in the branch tree, the step specifically includes: 1) ffl The above-mentioned hit test method determines which branch the user has selected 3⁄43⁄4^ _;

 2) In the middle of deleting ^ ffl should be

 3) Redraw the logical structure diagram.

 When adding and deleting branches, it is necessary to ensure the completeness of the branch tree. In addition, it can be added and deleted. It can only be the ELSE IF branch or the ELSE branch of the IF structure, the CASE branch of the SWITCH structure, and the DEFAULT branch. , ELSE branch or space division, there is one, in the SWITCH structure, DEFAULT branch and empty branch must also have one. The branch product corresponding to the loop structure cannot be added to the branch.

 Using the step of changing the wing, the step specifically includes:

1) ffl The aforementioned hit test method determines which branch the user has selected 3⁄43⁄4# _;

 2) Display a edit «, which shows branch judgments;

 3) The user modifies the branch decision in the β;

 4) Save the new branch decision to the 3⁄4Χ.

 The step of invoking the in-laws as a truncation code, the step specifically includes:

 1) The aforementioned hit test of the shed to determine which of the choices the user has selected;

 2) Display a code, if the statement: ^ like the existing code, see the code in the editor «

 3);

)

 The steps used to hide or display an image. For programs with more internal logic, its logical structure diagram will also be more complicated. In order to highlight the key points, it is easy to separate WfiiT, logical structure, or for JBiSff path, part of the logical structure diagram can be hidden. This example can be used to hide objects. When you draw a hidden 3⁄4] image, you can ignore it. You can also draw an alternative icon. Figure 14 is a diagram showing an illustration of a partially illustrated hidden object, as shown in Figure 14, if the hidden image is a branch tree or a statement block, a small rectangle 1405 (referred to as an alternative rectangle) is drawn. The small rectangle of 3± can select the object and resume normal display. If the hidden object is a branch, draw a dotted line 1406. Hidden difficulty a steel -, direct or indirect sub-carving while hiding. This step specifically includes:

Each field that adds a record status, the field is "normal" and "hidden", the default value is "normal"; to hide a certain biliary, set its state to "hidden", and redraw the logic ^ I Figure. When calculating the branch tree and 驹:^[the width and height of the image, check its state, if the state is "hidden", then its width and height are the width sio and height sii of the giant shape, and the sub is ignored. Calculate the branch - the width and height of the bribe, check its state, if the state is "hidden", ignore the child as the empty rubber to calculate the width and height. When drawing an X image,

If the state is "hidden", then only one cut giant is drawn, and the subimage is ignored. For the division 3⁄43⁄4, if the state is "β 急存", then only a 3⁄4 Λ point and an exit point are drawn a dotted line, but the branch draws , ignore the child.

 The step for generating a vine Mm code by a logic ,3⁄4图, the step specifically includes:

 Starting from the top layer of the rubber, recursively scan each (acorn, for the hesitation, directly output its corresponding code, for the rubber, if it is an empty branch, then ignore, otherwise «out branch judgment and branch role: W first micro as C++" { "or "begin" of PASCAL), then scan each object, and finally output the end of the branch scope (such as C++ "}" or PASCAL "end"). For branch tree objects, if

^A special branch tree of an embodiment»M, such as the _sw itch branch tree of C++, then the peripheral code such as switch(- - -) i starts with "{", scans each sub-part, and finally outputs the function. The end of the field is like " } "; if there is no branching tree, it will be 3 straight lines 3⁄43⁄4^3⁄4. When generating the code, you can add the balance as needed. In addition, each steel has a vine level, such as the level of the top layer image is ο, and the top layer is 3⁄4. The straight level is 1, m, which can be indented according to the «^ under calculation code. And add a space to

 Code that is used to typographical formatting.

 In addition, some of the diagrams can be used for some tricks. As shown in Fig. 14, hemp can be added to the left end of the top layer, and the drunk entry icon 1401 is added, the width and height are S7, and the right end is shown by the outlet icon 1402, which has a diameter of S8. In the setting system, the origin is shifted to the right by S7, leaving the health of the entrance icon.

 As shown in Fig. 14 j 3⁄4, the branch line 3⁄43⁄4 rectangle 1403 of the branch tree may also be referred to as an entrance rectangle, and the entrance rectangle is added to make the structure of the logical structure clear. When calculating the width of the branch tree, the width of the entrance rectangle on the force port S9 In the calculation of the coordinates of the entry point of the ^ branch, the offset of the abscissa of the giant shape i ^ is also added. When the branch tree is drawn, the entry line is drawn as a rectangle.

As shown in FIG. 14, when the code of the "I" block contains a return statement, the icon of the block object can be drawn as shown in FIG. 140, that is, a diameter of S10 is added at the right end. The round, 3⁄43⁄43⁄43⁄4 is also an exit for ». As in the previous male leg, the present invention can be used to design a drunk logic map (for example for detailed design), and can also be used to generate a logical structure map (for example for a tender section) from existing flaws. In this embodiment, a logic structure is generated according to the existing method, and the step 8 of the ship branch tree or the statement according to the Mm code «. In this embodiment, step B specifically includes: a logical structure of the code, the branch structure of the core code generates a corresponding branch tree size, and the i ^ block according to the structure of the heir code And the branch tree X Weikou leg inquiries: ^ Yu embedded «W should be level 3⁄4X rubber into the" ^ is described as follows:

 First, break the translation code into pieces. The fiber code piece refers to the code unit of the structure, and the code piece categories are: judgment, daring number, 驹 block, and the number is divided into domain start and domain end. A decision is a piece of code that contains a judgment S3⁄4S megabyte, which is used to control the execution or non-execution of its query, such as C++ if (...; ), dse 3⁄4..·), else, while( ...), switoh(...), caseN; Basic IF..., ELSE IF..., ELSE, DOWHILE~. The decision must be complete. For example, C++'s if(...) is a complete decision, no matter how long it is in the parentheses, it is part of the decision and cannot be split. The domain symbol is the code of the scope of the decision, the domain begins with the decision, the start of the scope of the specified decision, and the end of the domain is used to specify the end of the scope of the decision. The typical domain 3⁄4n c++ "{" and "" }", "begin" and "end" of PASCAL; Statement: ^ refers to the code sequence of jl drunk corresponding to the former user M, which cannot contain the judgment. Comments that are within a code slice can be ignored or used as part of the code slice. Comments outside the code slice can be ignored or attributed to subsequent code slices. 1C is an example of code slice classification, as shown in FIG. 1C, the code slices corresponding to codes 111, 113, and 11 are determined; the code slices corresponding to codes 112 and 117 are 3⁄4^, and the code slice corresponding to code 112 is 3⁄4. The code slice corresponding to the code 117 is the end of the field; the code piece corresponding to the code 114, 116, and 118 is the statement block. ―

 This male example deletes the code structure, the 'mi additional number U three fields of the face structure to store the code piece, the code piece guarantees such as the linked list

»Groups like «In the container. M rnt, for the judgment class code piece, it should be judged whether it is a branch tree, and whether the judgment belongs to a subsequent branch of a branch tree, and the keyword that can be separated from the decision is a special branch tree, such as C++ code. Switch » word; subsequent branches can also be identified from the determined keywords, such as for C++ code, if it is the decision of the else keyword or the decision of the elseif keyword, it belongs to the subsequent branch, if a decision class code is ^ The branch tree ffi continues to branch (both Ke can and 歹! D, it is recorded in the additional category.

 Sometimes, the 3⁄4ff number is omitted as 'M" + "code. When the scope has only one statement, the Shan number can be omitted. For example, in the Basic 3F...THE... structure, the balance is unbalanced.亍, the scope of the statement after the same line, ΊΉΕΝ is: I3 start, domain end omitted; IF...THEN...END IF structure, ΊΉΕΝ and END IF are: start and domain end, and IF ...THEN...ELSE IR..THEN...ELSE...END In the IF structure, the end of the field is omitted before ELSEIF, and the end of the field is omitted before ELSE. When the code is decomposed, it is guaranteed to have a matching start and end of the field after each decision. If ^ is omitted, the corresponding biliary code should be added, and the code is empty.

 The code also considers the implicit branch. When the IF structure lacks the ELSE branch, the SWITCH structure lacks the DEFAULT branch, and the loop structure that is considered to be a winged hook, it contains implicit branches. Three pieces of code should be added to the implicit branch: one decision, one field starts, one field ends, and their code is empty.

 Different programming languages, different syntax and keywords for code decomposition, the specific ¥ process can be used in the compilation of technical analysis and linguistic techniques to achieve real MM pre-technical skills are not detailed here <=

 After the completion of the generation, the further tree is inch. FIG. 15 is a schematic diagram of the flow of the Chengji film as a structure script in the embodiment. As shown in FIG. 15, 1501 is the input parent object, and the branch tree generated in the flow is # ft^J:^3⁄43⁄4ft3⁄4TO (in the image, The parent X ship can be a branch. 1502 is a code slice queue, the thief chip is intoxicated by the code, and is stored in a first-in-first-out queue, called a code slice queue. The "pop-up code slice" of the latter user refers to The first piece of code is taken from the code slice queue, and the code piece is deleted from the code slice queue; the "code piece" described later, when not limited, is a new piece of code that pops up; The HA temporary queue means that the 弋 弋 保存 保存 保存 保存 保存 保存 保存 保存 保存 保存 保存 。 。 。 。 。 本 本 本 本 本 本 本 本 本 本 本 本 本 本 本 本 HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA HA When calling, the top layer generated by step A is divided into 3^4 (働5^m, all pieces of code package ^ ®ί, when processing a branch, in order to further process the code piece corresponding to the branch, the branch is taken as : 53⁄43⁄4, the branch of the branch The corresponding code slice is called as the input code slice in the temporary code slice queue.

 Step 1503 determines whether the code slice is empty. If it is empty, it ends. Otherwise, step 1504 is executed. Step 1504 pops up the code slice. In step 1505, the number of the code slice is determined. If the number U is a statement block, step 1511 ' If the number lj is a decision, go to step 1521, such as the number (J is the number, because the 3⁄4 number should not appear at this time), perform step 1590, step 1590 reports! ^ Wu and exit.

Step 1511 generates a sentence: ^yu' and embedded in the middle, this embodiment is an auto-incrementing character in the oak shed» as the name of each sentence: ^ rubber, such as V, V, V. Step 1512 security code, copy of the code content of the tree chip Go to the code field of 3⁄4^, and then return to step 1503 to process the next code slice.

 Step 1521 generates a branch tree 3⁄4tm, and «to, in step 1522, according to the additional category of the code slice, the branch tree is a branch tree, if the branch tree, step 1531, otherwise step 1541. Step 1531 protects the code and copies the code content of the code slice to the peripheral code field of the special branch tree. Step 1532 determines whether the code slice queue is empty. If it is empty, then the terminal 1590 reports an error and exits, if no, the tracker ίϊ« 1533 pops up the code slice, and in step 1534, the number of the code slice is 1", if the class J is In the case of Gangtai, return to the operation of step 1532 (the beginning of the code is the beginning of the code, step 1572 ignores the domain of the peripheral code, if the number is deleted, then step 1541, if the category is not the beginning or the judgment, then step 1590 is reported and drop out.

 Step 1541 generates a split, in the branch tree sculpture generated in step 1521, the instance is in the form of an auto-incrementing word ^ as a name of the «~ rubber, such as 'a', V, V. Step 1542 saves the branch decision, and the code content of the weight chip is copied to the branch decision field of the sample. Step 1543 determines whether the code slice is empty. If it is empty, step 1590 is reported and exits. Otherwise, the code piece is popped up in step 1545, and in step 1546, the class of the code slice is determined. If the number ij is i3⁄4T, the shell 1 "Fix step 1547, otherwise step 1590 is reported and exits. Step 1547 «- is used to determine whether the 3⁄4 number matches »:^ and the value is assigned to a 3⁄45T. Step 1551 determines whether the code slice queue is empty. If it is empty, the IJ executes the error 1590 and exits. Otherwise, the execution is terminated. 1552 pops up the code slice, and determines the number ij of the code slice in step 1553. If the number U is: 3⁄4JF, then N is incremented by 1. If the domain is over, N is decremented by 1, and then it is judged whether the value of N is 0 or not. If it is 0, it indicates that the corresponding domain end has been popped up, domain matching is implemented, and step 1562 is performed. Otherwise, in step 1554, the chip is pushed into the temporary queue, and the process returns to step 1551 to process the next code slice. Step 1562 Process the branch code, use the split generated by step 1541' as 5i3⁄4, use the temporary queue as the code slice queue, enter the «call, and then clear the temporary queue.

 Step 1563 determines whether the queue is empty. If it is empty, it ends. Otherwise, the code slice is popped up in step 1571, and the number ij of the code slice is judged in step 1572. If the category is the end of the domain, the process returns to step 1571, if the category is determined. Step 1573 executes step 1573, otherwise returns to step 1505 to process the subsequent code slice until the end. Step 1573 determines whether the code slice belongs to a subsequent branch of the previous branch according to the dependency 1" of the code slice, and if so, executes step 1541; otherwise, the current branch tree has been processed, and returns to the processing chip 1505 to process the subsequent code piece until End.

 Another implementation of the step B of Fig. 6 3⁄4⁄4 is described in detail above. This example uses the Mlift code to generate a picture, which can be used to synchronize the structure of the bare ship with the code when writing the code. It can also be used to generate the logic structure chart during the test. The embodiment for the statistic » path is described in detail below, and based on the above-mentioned masculine case, includes a step for immersing the path. When the statistical path is used, the logical structure diagram of Hertz may not be drawn, that is, the steps of calculating the size and structure of each structure and drawing the illustration may be omitted. ·

 The path is the execution route, and the cost of the boat is the same: 驹Combination and domain division are combined, and the elimination is said to use 驹 combination and/or branch combination to trace the path. The branch combination can be used to count the branches of the path. The statement combination refers to the sequence of the i ^ block executed by the path, which consists of the statement block name according to the execution of the path; the branch combination refers to the sequence of the branch experienced by the path, which is composed of the branch name according to the path experience; 3⁄4 means , the statement group ^ own line of the statement, the branch group ^ recorded path experienced branch. We refer to the statement in the statement combination of the path as the execution statement of the path, and the branch in the branch combination of the path is called the experienced branch of the path. In addition, due to the need of statistical paths, the path is divided into two categories: normal and return, the number U is the return of a statement block with a return statement, so it ends early; the normal path is terminated at the end of ». When the path is counted, the default class of the path is normal. When a block has a statement block that returns a statement, the class of the path is set to return, and the path ends.

 The steps of the path ii ^ specifically include: the structure X rubber described later, for each input path set as the initial data, the input path of the path statistics top-level image division *K contains a path; for the statement image, in the input path set The name of the statement block is recorded in each path; for each point, the branch name is recorded in each path of the input path set, the output path of the previous ^@ is used as the output of the next child; for the branch tree The copy of the AS set is used as the set of transmissions for each image. The following describes the detailed steps of recursively calculating the statistical path of each structure during the steps:

 16 is a schematic flow chart of the statistical path in the embodiment, wherein FIG. 16A is a flow chart of the statistical path when the sample block is a sample block, and FIG. 16B is a flow chart of the statistical path when the branch is a branch, and FIG. 16C is a flow chart. Schematic diagram of the statistical path of a branch tree.

Fig. 16A is a flow chart showing the statistical path when the block is a statement block. As shown in Fig. 16A, step 1603 scans the path set, that is, scans the f-in path set 1601, and each path is determined. In step 160, its IJ is judged. If the category is returned, the leg returns to step 1603. A path, otherwise, in step 1605, the code that earned the query contains a return statement. If so, WMs 1606 sets the path of the 3⁄43⁄41" to return? Step 1607, otherwise, the straight g3⁄4 line is step 1607, and the step 1607 is recorded as the name of the block, that is, the block name is added at the end of the combination of the paths. After all the paths are processed, the output path set 1602 is obtained, and the number of paths in the output path is consistent with the input path set _D.

 Fig. 16B is a flow chart of the statistical path when the rubber is a branch. When the rubber is a branch, the rubber X rubber is calculated in turn, and the Itr-? ^ calculated output path set is taken as the lower After all the sub-3m4 are calculated, the output path of the sub-sum is the output path of the sub-picture. As shown in Fig. 16B, step 1611 records the branch, and the branch of each path of the 3⁄4SrA path set is added to the branch at the end. Step 1612 If the scan is completed or there is no sub-carving, then step 1613 is performed. Step 1613 is to determine the 3⁄4 sub-segment (if it is not a statement block, if yes, then step 1614, otherwise step 1615, step 1614 recursively perform Figure 16A The calculations shown, and the calculated set of paths as the set of input paths for the underlying data; step 1615 recursively performs the calculation shown in Figure 16C, and the resulting set of paths is used as the set of input λ3⁄4 paths for the lower image. Step 1614 and steps After execution of 1615, both return to step 1612 to process the next sub-X rubber until the end.

 Figure 16C is a flow chart showing the statistical path when X is married as a branch tree. As shown in Fig. 16C, in step 1623, the path is deleted: the path IJ of the input path set 1621 is copied to the output path 1622, and deleted from the AS path set 1621. Step 1624 scans the sub-pairs (divided, after a branch is broken, step 1625 takes the input path as a set of input paths for the branch. Step 1626 performs the calculation shown in FIG. 16B for the part, and then returns to the step. After 1624 processing, all branches are calculated. In step 1627, the output path of each branch is totaled in the output path 1622 of the pin branch tree, and the calculation ends.

 The input path of the layer 3⁄43⁄4^ contains only one empty path, which is the path of the combination of the sentence and the division and the combination of the paths; the top-level output router is the set of paths. Each path in the path set records the statement executed by the path: the sequence of the image and the sequence of the Luna calendar.

Figure 17 is a schematic diagram showing the path of the process. As shown in the logic diagram of FIG. 17A, there are three paths, and the combination of sentences is: ac, bc, c; as shown in the logical structure diagram shown in FIG. 17B, there are six paths, and the combination of sentences is: adc, Bdc, dc, aec, bec, ec _; as shown in the logic diagram of Figure 17C, there are 8 paths whose statement combinations are: adc, bdc, fdc, _g dc, aec, bee. fee. gec.

 When » is more complicated, the number of paths will be very large, and it is unrealistic to design a test to cover all paths. You can hide the object corresponding to a 3⁄4 code, and when traversing the path, ignore the hidden and its direct or indirect sub- 3⁄4, and reduce the number of paths by EEE. If you draw a logical structure diagram, you can select the object to hide by the user, otherwise you can hide the deep object.

 The example also includes the steps of drawing a path of the ® ί, drawing an illustration of a path at a time, the steps specifically including:

 Draw a diagram of the path as it passes through each branch. The branch that the scan path is going through, for each branch, if the branch has a parent object, the shell! is a branch tree, and the entry point of the branch line connecting the thick line of the embodiment and the entry point of the branch, 'and ^ The exit point and the exit point of the branch. '

 Draw a path through the various time-sharing diagrams. The branch that the scan path is going through, each branch of the needle, in this embodiment, the entry point of the sub-rubber, the sub-X rubber, and the exit point are connected by a thick line segment, that is, the entry point of the connection entry point and the first sub-point, ~T Henan exit point - the entry point of the second child object, ..., the exit point of the last child object and the exit point of the branch.

 Draw the path through each statement: ^ like the icon of the inch. The scan path is experienced, for each block, the 3⁄4⁄4 case is filled with a slash to fill the radius of the rectangle 3⁄4 ^ 3⁄4i the block.

 17A and 17B show the present embodiment in which a path is drawn, Fig. 17A shows the path be, and Fig. 17B shows the path ec. It can be seen from various embodiments that the method of the logical structure diagram proposed by the present invention can be designed in a fast ship, and can be bidirectionally converted between the code and the logical structure diagram, which can be used to count and show the program path. The program path statistical method proposed by the present invention can automatically count the Heheng path and draw a schematic diagram of the path. The above-mentioned embodiments are merely preferred embodiments of the present invention, and are intended to be illustrative, and not restrictive, and modifications, variations, or equivalents may It is within the scope of the invention.

Claims

Claim

1. A way to illustrate the logical structure. The curtain is in the process, including:

 A. Generate "top layer resolution;

B. Nesting branch tree wipes I or statements ^ m in any level of oak _;

 C. Calculate the various sizes and 443⁄4 and draw the illustration.

 2. The method according to claim 1, wherein the logical steel is divided into an office structure and a branch structure; the rubber is used to describe the sequence structure, and the branch tree is turned over to describe the branch structure. The sub-segment is a branch describing the top branch or branch structure; each household has at least two sub-insoles nested in the full branch.

 3. The method of claim 1, wherein the household includes: the range of being nested within at least one of its ranges of 3⁄4⁄4.

 4. A method of gff logical representation of a household according to claim 1, wherein the different branch structures are described by a refined branch tree.

5. Leakage and profit requirement 1 fiber's logical structure diagram method, which is characterized in that the step B branch tree carving and / or block is specified by the user.

 6. Leakage and profitability The 5-legged logical structure diagram method is characterized in that the β-difficult branch tree Xie/or 驹: ^The rubber's 3⁄4m is determined by the user t.

 7. The salary logic structure diagram method according to claim 1, wherein the step B leg branch tree/or the statement block is generated according to the code.

 8. The logic structure diagram method according to claim 7, wherein the step B specifically comprises: solving the ilfgJW side spoon logic structure 7 according to the branch structure of the translation code, the branch tree should be 3⁄4 inch The image is generated according to the grinding structure of the Wo's code, and the corresponding block 5 is generated (such as the image of the branch of the tree branch).

 9. An alcohol logic method according to any of claims 1 to 8 wherein the dimensions of the ship comprise: width and height of each pan.

 10. According to the method of 9 households, the gff logic ^ I diagram method, 3⁄4 is in, mc ^M:

 Ci. Starting from the top level, for each X marry, calculate the width of each sub and the height and height calculated according to the width and height of the sub, ^^ include:

 Calculate the width and height of the statement rubber according to the following formula:

Width = S1 _;

Height = S2 _;

 Where S1 is the width of the image and S2 is the height of the i-question;

 Calculate the width and height of the rubber according to the following formula:

 Width = S1 + 2*S6 (N-0);

 Width =∑W + ( +1)*S6 (N>0);

 Height = S3 + S4 (N = 0);

 Height = Hm + S3 + S4 (N>0);

 Where 'N is the number of sub-Shes of the 3⁄4 rubber, S1 is the width of the 驹 rubber, S6 is the distance between the left 3⁄4^ to the connection, ∑W is the sum of the widths of all the sub-objects, and S3 is the branch The height of the note, S4 is the distance between the branch main and the branch, and Hm is the height of the child X image with the highest height;

 Calculate the face height of the branch tree bile according to the following formula:

 Width = Wm;

 Height = ∑H + (N-1) * S5;

 Where Wm is the width of the widest sub-X rubber, ΣΗ is the sum of the heights of all sub-objects, N is the sub-number (the number of rubbers, and S5 is the distance between the two branches).

 11. Leakage and profit requirement 10 繊 禾 逻辑 逻辑 逻辑 图示 图示 , , , 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 图示 图示 X X X

12. The method of the present invention as claimed in claim 11, wherein the step C1 further comprises: Calculate the male height of each Weng;

 Calculate the branch line height of the branch tree according to the following formula:

 Entrance line height = Βι + (Ηη- Βη) + Σ(Η2..Ήη-ι) + (N-1)*S5;

 Where Bl is the height of the first sub-object, Hn is the height of the last sub-object, Bn is the height of the base of the last sub-object, and Σ(Η2···Ηα-ι) is other than the X image of the D^l** The sum of the heights of the children, N is the sub-S5 is the J? giant distance between the two ^ 53⁄4.

13. The method according to claim 12, wherein the tree is in the height of the 3⁄4m of the i-th to include: from the top layer 3⁄4 (starting with the oak, calculating each child for each recursion first) The base height of the X rubber is ₇ and the height of the X rubber is calculated according to the height of each sub-paste.

 Calculate the height of the rubber according to the following formula:

鹓 height = S22 _;

 Where S2 is the height of the statement;

 Calculate the height of the vine of the S« according to the following formula:

Male height = 0 (N = 0) _;

 Male height = Bm ( >0);

 Where N is the number of sub-images and Bm is the height of the sub-object with the highest height;

 Calculate the height of the branch tree carving according to the following formula:

Hertz height = [Βΐ + (Ηη- Βη) + Σ(Η2...Ηη-ι) + (N-1)*S5] /2 + Bn _;

 Among them, Bl is the second (the height of the oak, Hn is the height of the ft*, the height of the Bn is the height of the rubber ring of the **3, and the Σ(Η2··.Ηη-ι) is the division of the ^^3⁄4^3⁄4** The sum of the heights of the children other than 3⁄4, Ν is the child X image | 3⁄4, S5 is the large distance between the two.

 14. The method according to claims 1 to 8 要求 ⁄ 要求 要求 户 户 户 户 户 , , , , , , , , 逻辑 逻辑 逻辑 逻辑 逻辑 逻辑 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示 图示

 15. The logic diagram method according to claim 14,, Xiao is in, 継Step C:

 C2. Starting from the top 3⁄4 rubber, 'for the oak, first calculate the entrance point coordinates of the rubber, and then calculate the entry point coordinates of each X image according to the recursion, including:

 According to the following formula, the coordinates of the entrance point are: ' '

 X=0;

 Y=H - B;

 Where X is the abscissa of the entry point, positive to the right, Y is the entry point mark, the direction is positive to the down, H is) (like the height, B is the height of the key); the Nth of the 3⁄43⁄4# is calculated according to the following formula Oak entry point coordinates:

Xn= ∑(Wi...Wn-i) + N*S6 _; where, for the child 3⁄4^1 starting from 1, 3⁄41 is the abscissa of the entry point of the crane ^^ ^3⁄4椽, positive to the right, ∑(Wi...Wn-i) is the sum of the widths of the children before 3⁄4m N ^3⁄4 ^ , S6 is the distance between the two 3⁄4 ^ of the left connection, and Υη is the entry point of the third image of the household 3⁄4 The 11⁄2 mark, down to positive, is the sub-entry mark;

 Calculate the branch point of the branch tree by the following formula?!

 Xn = 3⁄4

 Υη = Υι - E/2 ( =l);

 Υη = Υη-ι + Βη-ι + Ηη - Bn + S5 ( >1);

 Where N is the sub-nickname starting from 1, 3⁄41 is the abscissa of the entry point of the image, the right is positive, Xi is the abscissa of the entry point of the branch tree object, Υιι is the first Ν ·^ΡΧ The ordinate of the entry point, down is positive, Yn-i is the ordinate of the entry point of the previous sub-object, Bn-i is the front] 3⁄4j3⁄4 height, Hn is the impurity N (the height of the oak, Bn is the woven N ^ image From the height, S5 is the distance between the two ^^ branches.

 16. The logic diagram method of claim 1 to S i-claim, wherein the representation of the full branch tree comprises: an entry line and an exit line.

17. The @ί logical structure diagram method according to any one of claims 1 to 8, wherein the graphical representation of the word "comprising" includes: stipulation.

18. The method according to claim 17, wherein the figure is: the household; the map of the rubber includes: branch mark ₀ ―

 19. The illustrated method according to claim 18, wherein the graph of the household image is included: the main text of the branch.

 20. The method according to any one of claims 1 to 8; wherein the graphical representation of the household; ^ comprises: a range rectangle. ...―.

 21; The method for graphically illustrating the logical structure of any of claims 1 to S, wherein the household step C comprises the steps of:

C3. From the top level, 3⁄4 (starting with Henan, first draw a 3⁄4 image for each Liu Xiang, and then recursively draw a picture of each sub-image.

 22. The method of the cake logic according to claim 21, wherein the step C3 further comprises: ignoring the hidden y mm .

 23 According to the claims 1 to 8 ^ 3⁄4, ifM's logic logic diagram method, characterized in that: "^ includes: a step for adding a branch in the branch tree earning.

 A method according to any one of claims 1 to 8 to claim wherein the method comprises the steps of: deleting the selected branch in the branch tree.

 25. A method according to any one of claims 1 to 8, wherein the tree is in progress, further comprising: a step of modifying the branch decision.

 26. The method of logic according to any of claims 1 to 8, the tree being in the step of: further comprising: a step of coding the code of the rubber.

 The method for illustrating a logic structure of a program according to any one of claims 1 to 8, further comprising: a step for hiding a port and/or displaying a sadness.

 28. The method of claim 1 to claim 3, wherein the logic is illustrated by the method, and the method is further comprising: a step for generating a weight from the logical structure diagram. "

 •

29. A claim 1 ^3⁄4 ^ logical structure diagram method method ^ path method, characterized by the step of using the system ffif path to step C. '

 - .

30. A method of statistically traveling a path according to claim 29, the tree being in, dm^: combination and/or branch combination. -

31. A method of statistically paying a payroll path according to claim 29, wherein the step of: said step of arranging comprises: rearranging said pulses for each of said sin-in path sets as initial line path statistics, The top layer of the 3⁄4« set contains only one final path.

 32. The method of claim 31, wherein the step of «statistics® ί path comprises: recording the block name of each statement for each path of the statement: fe3⁄4^, 3⁄43⁄4Λδ; S3⁄4), the front output path set is used as the lower path set of the rubber; for the branch tree pulse, the copy of the input path set is used as the input path set of each "^".

 33. The method of claim 32, wherein the step of calculating the path of the boat is included in the step of arranging the path of the drunken path. The branch name is recorded in each path of the pair of fibers.

 34. The method of claim 3 to 4, wherein the step of the household path comprises: when the path contains a statement that returns a statement, the path ends.

 35. A method according to any one of claims 31 to 33, wherein the step of the step of the «system^» path comprises: ignoring the hidden X-like and hidden direct or indirect Child X.

 36. A method of routing a ship according to claim 29, wherein the step of including the leg of claim 1 Co

37. A method of statistic ® path according to claim 36, wherein the step of drawing a path of ® ί is included.

 38. A method of routing a leg according to claim 37, wherein the step of drawing a path of the leg comprises: drawing an icon when the path is a tree object;

 Draw a diagram of the path through each branch 3⁄4 ^; - Draw a path of ± ^ for each icon.