JPS61289380A - Document generator - Google Patents

Abstract

(57) [Summary] This bulletin contains application data before electronic filing, so abstract data is not recorded.

Description

DETAILED DESCRIPTION OF THE INVENTION [Technical Field of the Invention] The present invention relates to a character creation device, and more particularly to a device for creating characters to be supplied to an output device such as a dot/printer or a dot display.

[Prior art 1 Dot printer using a character generator (e.g.
When printing on a thermal transfer printer or the like, character pattern data of the same pattern as the font to be printed is generally stored in each character storage area of the character generator.

[Problems with the prior art] For this reason, character (1) ELITE as Font
, PICA, GOTHIC, etc. If you try to use a slanted font such as ITALIC, you will need a larger storage area for each character generator than that for ELITE. Inconveniences arise, such as the need to increase the capacity of the generator.

[Object of the Invention] The present invention solves the above-mentioned problems, and without increasing the capacity of the character generator, in other words, by efficiently using the character generator, it is possible to create characters ( For example, the present invention aims to provide a character creation device for creating slanted characters such as ITALIC.

[Summary of the Invention] According to the present invention, character pattern data is stored in each storage area for one character of a character generator in the form of a decomposed pattern (font) of an output character to be supplied to an output device such as a printer, and (output If you try to insert the same pattern as the character pattern, there will be a part that protrudes from the storage area for - characters, so this protruding part is placed in the empty area of the same storage area.) When reading The gist of the present invention is that out of the above character pattern data, the data placed in the empty area and the remaining data are combined and output to create a character.

[Example] Figure 4 shows character creation according to the present invention! 1 is a block diagram of the application of 1tW1 to a word processor. The illustrated device consists of two parts, 1. R.A.
M2. The ROM 3, the print controller 41 that controls the print unit 5, the display controller 6 that controls the display unit 7, and the CPU 8 that controls these items, and the RAM
2. ROM3 and CPU8 are interconnected via an address bus. In addition to a normal microprogram area, the ROM 3 is provided with a character generator 3a for storing character pattern data. In addition to the usual area for storing various calculation results, RAM2 has a key area 1.
Document memory 2 that stores document data that has been stored for a long time via
A print buffer (FB) 2b is provided which receives and outputs data in the form of an output character pattern by combining data from the character generator and the character generator (details will be described later).

When printing data in one document memory 2a, the word processor shown in FIG. The data of the output character pattern is read out and synthesized into an output character pattern on RAMZ from the print buffer (PB) 2b by the print controller 4. Alternatively, it is provided with a function (character creation function) to send it to the display controller 6.

Next, the pattern of character data contained in each storage area (for one character) of the character generator 3a will be explained in comparison with the pattern of the output character output from the buffer (PB) 2a.

- An example is shown in FIG. 1. The lower part (B) of FIG. 1 illustrates character patterns stored in three character storage areas of the character pattern generator for one character M, A, and N. The upper part of FIG. As you can see from the comparison of the two patterns, for example, for M and N, the upper right part is the upper left area of the corresponding character storage area of the character generator. It's in. The pattern in this portion is connected to the patterns in the remaining areas B and F during printing. In this embodiment, the pattern in area 3 is printed at the upper left of the digit of the next character (for example, character A). As can be seen from the above example, the character generator 3a generally handles output character patterns that are too large to fit into the character storage area.
It is disassembled so that it fits within the same storage area. In other words, a certain fixed width is here 24
This effectively uses the storage area for one character, which is only available in the 24-bit The empty area (area ■ shown in FIG. 1) is used to place it there.

Next, a device for creating output characters from a character pattern on the character generator side as illustrated in FIG. 1 will be described. Although this can be realized in various ways, a first embodiment will be described with reference to FIGS. 2 and 3.

Figure 2 shows the main parts of the character creation device.

(a) is the storage area for one character of the character generator (hereinafter referred to as one character CG) in RAM2 (11th ffl
) corresponds to the data code of the character to be printed in the text memory and represents one character CG specified by address specification. The data read from is output as an output character pattern. 1 character CG
has a size of 24 x 24 bits, and area ■ is 1~
It occupies 4 columns x 1 to 12 rows, and the portion that cannot be accommodated in one character CG with the same pattern as the output character is stored here (see FIG. 1(b)).

Area ■ occupies 1st to 4th columns x 13th to 24th rows of one character CG, and area [F] occupies 5th to 24th columns x 1st to 24th rows. The print buffer PB has a print area of 1 to 24 columns x 1 to 24 lines, and a save area C that occupies 25 to 28 columns x 1 to 12 lines. In the save area C, the data in the 1-character CG area ■ is taken and saved once, and then the data in the 1-character CG area ■ is saved.
After the data in areas ① and [F] are read out, it is used to transfer the data to area 0, which occupies 1-4 columns x 1-12 rows of the print buffer. Area ■ occupies columns 1 to 4 and rows 13 to 24 of the print buffer,
This is an area for receiving data in area (■) of one character CG. Area @ is 5 to 24 columns of printer buffer x 1 to 2
This area occupies four lines and receives data read from area [F] of one character CG.

Let's briefly explain the procedure for synthesizing one character (for example, M) with reference to Figure 2. One character CG on the current (L) side
Assume that the letter M is stored in a pattern as shown in FIG. 1(b). Also, the evacuation area C is empty (
all "0"), first transfer the contents of area C to area C as shown by ■.0 Next, transfer one character CG as shown by ■
Move the protruding part of the letter M in the area ■ to the evacuation area C 6. Next, move the area of one character CG as shown by ■.
Move the M part in area [F] of print buffer 1PH to area [F] of print buffer 1PH.0 Next, move the M part in area [F] of 1 character CG to area ◎ of print buffer, as shown by ■. Here, in the print area (D+E+G) of the print buffer.

Among the M patterns shown in Figure 1 (B), (one letter C of M
Among the contents of G) area ■ and [F] contain the same pattern (patterns other than the patterns in the protruding part). Therefore, give the output command as shown in ■ and print it. Output the data in the area. Finally, as shown by ■, the 0th order output command moves the protruding part of the character M in the save area C (that is, the data that was in the area ■ of one character CG for the character M) to the print area area ■. As a result, the protruding portion of the letter M in area 2 of the print area is output. In this way, the previously outputted M part and the protruding part are combined and combined to create an output character pattern of 1M.

In accordance with the procedure for synthesizing one character briefly explained above, refer to the flowchart shown in Fig. 3 for the operation when actually printing on a line according to the data code stored in the text memory 2a of the RAM 2. (■ in the diagram)
, ■, etc. correspond to ■■, etc. in the above-mentioned procedure) First, with the printer set at the starting position of the print line, data of the first character to be printed stored in the one-sentence memory 2a is Determine the address of one character CG of the character generator 3a corresponding to the code (step 1
0), and in step 11, the 12-bit data (12-bit data in the 25th column to the 12th row of the print bus sofa PB)
1 of the 1st column of the sofa PB
- Move to row 12 (column 1 of area O). In step 12, the 12-bit data in rows 1 to 12 of column 1 of the addressed single character CG (data in column 1 of area ■)
is moved to the 25th column of print buffer PB, rows 1-12 (1st column of area ◎), and in step 13, 1 of 1 character CG is
The 12-bit data in the 13th to 24th rows of the column (area
Repeat one or more operations for moving the data in the first column to the 13th to 24th rows (first column of area ■) of the print buffer PH for the remaining three columns. As a transfer operation, move the data in the second column of area ◎ to the second column of area ■ Step 14) Move the second column of area ■ to the second column of area ◎ Step 15) 2 of area ■
Move the column to the second column of area ■ (step 16), the third
As a column transfer operation, move the data in the third column of area ◎ to the third column of area ■ in step 17), 3 of area ■.
Move the column to the 3rd column of area ◎ (step 18), move the 3rd column of area ■ to the 3rd column of area ■ (step 19)
), as a transfer operation for the fourth column, move the fourth column of area ◎ to the fourth column of area ■ (step 20).

Move the 4th column of area ■ to the 4th column of area ■ Step 21) Move the 4th column of area ■ to the 4th column of area ■ (Step 22) By performing 0 or more operations, print buffer C
The data for the unprinted part of the previous character that was in the area (here, there is no previous character, so all "O" empty data) is moved to area O, and the data in area ■ of the current character CG is transferred to the print buffer area. This means that it has been moved to 0.

Next, in step 23, the value n of the column pointer to be transferred is set to 5.
In step 24, data in the 5th column of the 1-character CG (data in the 1st column in area [F]) is transferred to the 5th column in the print buffer FB, and in step 25, the value of the pointer is incremented. In step 26, it is determined whether the incremented value n is greater than 24 or not. Here n
Since x6<24, only 1. By repeating the above operation and transferring each column up to the 24th column, all the data in the area [F] of the current character CG will be transferred to the area ◎ of the print buffer PH. Here, the contents of areas ■ and [F] of one character CG are placed in areas ■ and ◎ in the print buffer PR, and the data for the unprinted portion of the previous character that was in area ◎ (in this case, none) is placed in the area. The printer is now ready for printing. When the transfer is completed, the pointer value n-25>24, so the process proceeds to step 27, and the pointer value n=1 is set. Then, in step 28, the data in the first column indicated by the pointer is output from the print buffer PB and printed. Increment the value n of the pointer (step 29), and repeat the above operation for the next column data until the incremented value becomes greater than 24, thereby outputting the data in columns 1 to 24 of the print area of the print buffer. Print. As a result, the portions of the first character (for example, M) located in areas 3 and [F] of the character CG are printed. When printing is completed, the value of the pointer becomes n=25>24, so the process advances to step 31. In step 31, it is determined whether the next code in the text memory is a command such as carriage return CR.

If the next code is print data, repeat the above operation for the next character (for example, A).

In this case, due to the completion of the previous operation, the print buffer F
Please note that the save area C of B contains the protruding part of the data of the previous character (first character) M that is in the area ■ of the first character CG for M. Therefore, in this operation, In steps 11.14.17.20, the data in the 1st, 2nd, 3rd, and 4th columns (the protruding parts of the letter M) in the save area C are moved to the corresponding columns in area ■ in the print area. It will be. Therefore, when step 22 in the current operation flow is completed, the protruding part of the previous character M is placed in area 0 of the print area, and the areas ■ and ◎ are filled with areas ■ and ■ of the one character CG for the current character A.
The parts in each will be included. As a result, when the printing in step 28 is completed for all columns, the letters M and A are printed on the paper as shown in FIG. 1(a).

In this way, character composition and printing processing proceed, and in step 31,
If the code in sentence memory is determined to be a command (for example, carriage return CR), the last character (
For example, the command is executed after printing the unprinted portion of N).

That is, in step 32, the 12-bit data in rows 1 to 12 of the 25th column of the print buffer γFB (the data in the 1st column of the unprinted portion of N in the area ◎) is transferred to the 1 in the 1st column of the print buffer FB. In step 33, clear the first column of the print buffer, rows 13 to 24 (first column of area ■), and do the same to the second column of area ◎. Clear the second column of area [F] Step 35) Move the third column of area ◎ to the third column of area 0 Step 36) Clear the third column of ■ Step 37), move the fourth column of area ◎ to the fourth column of area ■ Step 38), clear the fourth column of area ■ (step 39) 0 then move area ◎ Clear step 40) and set the pointer to 5 (step 41)
Then, area ◎ is cleared one column at a time from the left column (step 42), and the pointer is incremented (step 43) to clear all area ◎. Here, in the print buffer, only the area ◎ contains the unprinted portion of the last character N, and the other areas are in a cleared state.

Then, in step 44, the pointer n-25>24 is determined, so the process proceeds to step 45, where the pointer is set to 1 to prepare for printing, and in step 46, the print buffer PB is output and printed one column at a time. The pointer is incremented (step 47), and in step 48 printing is performed from the first column to the 24th column until the pointer n exceeds 24.

As a result, the part where the final character N is stamped (the part in area 3 in FIG. 1) is printed. As a result, the value of the pointer becomes nm25, so the process proceeds to step 49, executes the carriage return command, and terminates the line printing.

FIG. 5 shows the main parts of a modified example of the character creation device.
The difference from the above (Fig. 2) is that there is a print buffer FB for one column (one dot) and (column 1 to row 24),
Same < Evacuation buffer C in RAM (columns 1 to 4 X1-12
The procedure for 0 synthesis, which basically uses the same synthesis method as the above-mentioned method explained in connection with FIG. 3, is shown by numbers ■ to ■ in FIG. be.

The details will be explained with reference to FIG.

First, in step 50, the value n of the pointer is incremented from O to 1, and a save buffer C is used to store the unprinted portion of the previous character.
Move the 1st column of the current character to the 1st to 12th rows of the print buffer PR (step 51), and move the 1st column of the current character CG area ■ to the 1st column of the save buffer C (step 52). ), move the 1st column of the 1-character CG area ■ to the 13th to 24th lines of the print buffer PB (step 53), and at this point, the print buffer is moved to the position of the 1st to 12th lines in the 1st line of the save buffer. The first part of the cancellation data of the previous character is contained, and the data of the first column of the 1-character CG area ■ of the current character is contained in the positions of lines 13 to 24. Contains the data in the first column of area ■ to be printed (when printing the next character). Therefore, in step 54, when the column number indicated by the pointer is 4, an operation of 0 or more to output and print the contents of the print buffer T is performed.
Repeat for the 2nd, 3rd, and 4th columns until .The result is that the unprinted part of the previous character in the area ◎ and part of the data of the current character in the The data for the protruding part of the current character being typed out is stored.

When 4 columns have been filled, the pointer n=4, and step 55
The result of the determination is YES, and the process proceeds to step 56, where a subroutine consisting of steps 56 to 61 is entered. This subroutine sends the data in area [F] of one character CG to the print buffer PB one column at a time starting from the left column, and
It performs printing processing, and in step 56 the pointer is incremented, starting at 5 and ending at 24).In step 57, the data in the column indicated by the pointer is read from the single character CG and transferred to the print buffer, and in step 58. The column data is output from print buffer γ and printed, and the column printed in step 59 is in the 1-character CG area [F].
It is determined whether it is the last column 24 of . When the last column is printed (meaning that printing of the current character is completed except for the unprinted portion), the pointer n becomes 24, the pointer n is cleared in step 60, and the next code in the text memory is entered in step 61. Determine whether is a print code or a command. If it is a print code, repeat the above operations for this new character; if it is a command, clear the data in lines 13-24 of the print buffer in step 62 and wait for the remainder of the previous character to be printed. The subroutine consisting of steps 63 to 66 prevents
The data in the previous character's 1 character CG area ■, which is currently being moved to the save buffer C (previous character's dying data), is printed one column at a time from the left side of the buffer. This is for sending to a line for output and printing. That is, in step 63, the pointer specifying the column to be read from the save buffer is incremented (initially from 0 to 1), and in step 64, another data indicated by the pointer is retrieved from save buffer C, and it is read out from print buffer 1 to 1. 12 row, and in step 65 the data of that column is output from the print buffer and printed, and in step 66 whether the pointer n=4 or not,
That is, it is determined whether all the unprinted data of the previous character in the save buffer C has been printed. After printing is completed, pointer n is cleared in step 67 and step 68
Run the command and complete the process.

[Effects of the Invention] In the embodiment described above as an example, the output characters are:
By using slanted characters, the slanted characters cannot be stored in each storage area of the character generator in their original form, so by placing the protruding part in the upper left area of the storage area, it will fit in the same storage area. However, the characters that can be handled by this invention are not limited to slanted characters; any characters that cannot be stored in their original form but can be stored by disassembling and arranging them can be used. , can be synthesized by this invention. In addition, in the example, the protruding part is placed in the "top left" area ■ of the storage area, but this is not limited to this, and in general, each character piece is placed within the storage area so that the disassembled character pieces do not come into contact with each other. It is sufficient to arrange the characters in different areas of the RAM, and the separated characters can be combined on the RAM side and output from the print buffer in the form of output characters.

As can be seen from the above, according to the present invention, character generators can be used efficiently and at low cost, and in particular, slanted characters can be printed on thermal transfer printers, etc. with a character generator of the same capacity as conventional ones. has.

[Brief explanation of the drawing]

Fig. 1 is a diagram showing an example of a character pattern stored in a character generator along with a character pattern of appearance according to the present invention, Fig. 2 is a diagram showing main parts of a character creation device, and Fig. 3 is according to Fig. 2. . A flowchart of character creation, FIG. 4 is a block diagram when the present invention is applied to a word processor, FIG. 5 is a diagram showing the main part of character creation *MIy) on the transformation side, and FIG.
6 is a flowchart of character creation according to FIG. 5. 2...RAM, 3...ROM, 8...
...CPU, CG...Character generator, PR...Print buffer, ◎...
・・Evacuation area, ■・Old ・・Data area to be evacuated. Patent Applicant Casio Computer Co., Ltd. Agent Patent Attorney Machi 111 Shun Tadashi R.
Method) 6. October 2, 1985
(1) Mr. Michibu Uga, Commissioner of the Patent Office, 7.1, Showa 6
0 Year Patent Application No. 131379 2, Name of Invention Expenditure Creation Device 3, Relationship with the Amendment Person Case Patent Applicant Address 2-6-1 Nishi-Shinjuku, Shinjuku-ku, Tokyo Name (14)
4) Casio Computer Co., Ltd. Representative Kashio
Tadao 4, Agent address: 1-13-4-5 Nishi-Shimbashi, Minato-ku, Tokyo Date of amendment order: September 24, 1985 (shipment date) Drawings subject to amendment. Contents of correction

Claims (1)

[Claims] When character pattern data is stored one character at a time in each storage area, if part of it exceeds the storage area, the data for the excess part is placed and stored in an empty area of the same storage area. character generator and
A read control means for specifying character pattern data from the character generator and reading it out one character at a time; A character creation device characterized by having an output means for combining arranged data and remaining data and outputting the result.