JPS59224893A - Data display method - 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 (1) Field of Industrial Application The present invention relates to a data display method in a data processing system equipped with a display device such as a CRT display. After storing the data in memory and reading it out, the original data is restored and displayed. This paper proposes a data display method that can display a large amount of information with α capacity.

(2) Prior Art In data processing systems such as those mentioned above, such as so-called microcomputer systems, video I? It uses the Rusk Scan method using AM (image memory). data stored in video RAM,
For example, is the character code CI? CI? is converted into a display signal by a display control circuit such as a T controller. The image will be sent to a display device such as a T-display, where it will be displayed as an image. In addition, the character code is generally external 1? It is stored in a non-resident area such as OM, and is read out as needed to display the video I? Stored in AM.

By the way, in general, when displaying information such as that written in books, for example, the contents of a cooking text are displayed using CI?
For systems that display on a T-display,
The number of character codes that must be left unused is enormous, and therefore a large area 1a is required to store the character codes.

(3) Purpose The present invention has been made in view of the above circumstances, and it compresses and stores display data, and restores and reproduces it when displaying, thereby reducing storage capacity and saving a large amount of data. An object of the present invention is to provide a data display method that enables information display.

(4) Structure of the Invention The data display method according to the present invention is a data display method for displaying character information stored in an image memory on a display device. The character information to be displayed is stored in the squared area of the storage device in association with data regarding the starting address of each of the 11 words in the specific area and the length to each car 8. A string of words is compressed and stored as a string of data in which each word is to be specified in the other area. The present invention is characterized in that the required character information is restored and reproduced based on the r< information and is stored in the image memory.

(5) Examples -1-1 The present invention will be explained in terms of monthly salary based on drawings showing examples thereof.

FIG. 1 is an implied external view of a display device according to the present invention,
Is this device a CI? Display section 1 consisting of T display etc.
0, a control section 20 that houses a microcomputer system, and an input section 30 that includes a keyboard and the like.

Figure 2 is a block diagram showing the circuit configuration of this device. 1 is a microprocessor that serves as the control center, 2 is a resident storage area consisting of system programs and data + I20M for each delivery, 34 years old 1? 4 is a non-resident storage area such as an EPROM, and 5 is a memory access controller for controlling data transfer between these areas and the display control circuit 7. 1 - a transfer control circuit consisting of rollers, etc.; 6, a CRT display device; 74 years old; converts display data stored in a video RAM (hereinafter referred to as VR/IM) area 31 allocated in work area J53 into a video signal; For CI?
This is a display control circuit consisting of a T controller and the like.

The data to be displayed is compressed and stored in the non-resident storage area 4. When positional data in the stored data is to be displayed, the data in the non-resident area a 4 is transferred to the display (hereinafter referred to as DSP) area 32 of the work area 3 and stored there. Next, the dictionary in resident storage area 2 (hereinafter D
ICT) area 21, pointer mark 1a (hereinafter referred to as PIN)
T) area 22 and word length ↑1g (hereinafter referred to as LHNG)
The data in the DSP area 32 is decoded based on the data in the area 23 (namely, restored and reproduced) and expanded and stored in the VRAM area 31. The data in the VRAM area 31 is then converted into a video signal by the display control circuit 7 and displayed on the CRT display device 6.

FIG. 3 shows work area 3 (7) DSP area 32. VRA
M area 31 and DTCT area 21 of resident storage area 2, P
2 is a conceptual diagram showing the data structure of an INT area 22 and a LENG area 23. FIG.

First, to explain the DTCT area 21, it stores all the words extracted from target display information, such as cooking text, and combines the corresponding code strings into a one-dimensional array, as shown in the figure. In the example, from dx address to dx+5 address, liver gate, E (dx4 address) GG
is stored. Note that alphanumeric characters such as h4.1 in parentheses indicate the ASfj1 representation of the corresponding character.

The PTNT area 22 stores the beginning addresses of many words arranged one-dimensionally in the old CTWi area 21, and in this example, dX of the beginning address of APPLE and IE
+4 is shown in d of the initial address of GG.

The length of the word stored in the old CT area 21 is stored in the LENG area 23, and here 5 is shown for APPLE and 3 is shown for EGG. In this LENG area 23, corresponding seeds for numbers and symbols (special characters) such as "*" are stored as they are.

This includes the DICT area 21, PINT area 22 and LHN
Although it is efficient to store t1 word information using the G area 23, it is wasteful to store single characters such as numbers and * in this manner. Therefore, 7!1 addresses, ie, Q-1/-J addresses, are allocated for these.

When the target is general display information such as cooking text that does not use various symbols, 1l=40 culm degree is sufficient. The data length of each area is 1 hide for the DICT area 21, 2 hides for the I''TNT area 22, and 1 hide for the ENG area 23.

Now, the nsp area 32 compresses and stores one byte per magazine, and the stored data is the address of the LENG area 23. In other words, the data at this address is the identifier for each word. And the memory of EGG * 8 PPLIE is Ak+1, I! becomes -1, Ilk,
What originally required 11 hides can be compressed into 1a with 3 hides, so the microprocessor 1 is made to continuously output this memory content, and Vl? In the quiet area 31, as shown in FIG.
It can be stored in the restored form as GG*APPLE.

So 1. Since the FNG area 23 stores one hide of data, it is impossible to compress and store more than 28-44 Q1 words. In other words, the total number of words is 256-
I17! If the number exceeds 1 unit, 1 high 1~
is impossible to express. Therefore, in this case L E N
A plurality of blocks are provided in the G area, and a code for identifying each block is set. In order to save storage space, the first block does not require the addition of this M'll-specific code, and codes for determining other blocks are not used in each block. In Figure 4, the LPNG area is 1, PNG],
FIG. 3 is a conceptual diagram of the data structure of each LENG area and the DSP area 32 when there are three blocks, LHIIG2 and LENG3. The discrimination code indicating that the second block is LPNG 2 is h''FE'', and the discrimination code indicating that the third block is LIiNG3 is h''FF.
”.

Therefore, both of these codes, LENG 1, cannot be used as codes to specify words.
In LPNG 1, addresses 7!12 to 253 are the storage area for word length information, and in LENG2. 3, there is no such restriction, so if fifi=40, in the case of 3 blocks, 726 +l1il (=256
3-40-2) word definitions are possible. As mentioned above, the total number of words is 256-1! , 7! , one Chinese word can be expressed in 2 bytes or less, and the first L
By storing words that appear frequently in the ENG area, the actual compression ratio can be increased.

Note that the stored content of the DSr' area 32 illustrated in FIG. 4 is 1.
, ENG 1 h “FO” address word, LENG 3
h word at address “8B” and h “0” at LENG 2
It is the 1st word.

Next, following the data creation procedure as described above, ept=40,
An explanation will be given by taking as an example the case of three blocks in the LENG area.

[Step 1] Extract all words that appear in the DTCT area 2 area 2,000 cooking text without duplication Step 2: Extract suffixes that can be shared among the word combinations selected in step 1, Select a combination of words with the maximum length of the suffix and/or prefix, and the length of the first letter, and connect them.

For example, E of AI'l'LH and E of EGG can be shared, and are connected to form AI'Pl and l1GG.

Hand jll'r3: Repeat step 2 for all ffi words to create a one-dimensional array structure called DIC.7IF. Obtain the T area.

[Hand B1αH] LENGV'I area (LENG1~1
, ENG 3) and r'lNT region creation procedure Step 1: For the 214 words with high frequency of appearance, I'l
The first address and length of each word in the ICT area are PIN TfrR area and LENG area LE.
NG 1 Store each.

Hand h1σ2: For the 256 words with the second highest frequency of appearance after the 214 words, similarly store the start address and word length of each word in the PINT area and LENG area LENG2.

Step 3: For the remaining words, enter the starting address and h f of each word.
l'j word length into PINT area and LEN (JJ[LENG
Each is stored in 3.

Step 4: LI'. Codes such as numbers and symbols are stored in addresses O to 39 of NG1.

0 When storing data in the PINT area, follow the steps below.

■ 1. If the address of ENG 1 is i (≧1e), then 1'1NT ((i-pl *2)←start address lower PINT((+-642)*2+1)-star)? Address upper ■ If the address of LENG2 is i, PTNT (256+ (+ -12β)*2) - Star 1 - Address lower PINT (256+ (i-12Il*2+]) -
Start address - high ■ If the address of LENG 3 is i, PTNT (5
12+i-j! *2) ← Start address lower PINT (512+1-17ri *2+1) - Start address upper [Procedure ■] Display data compression procedure The display data is compressed to 1 or 2 bytes per word as described above. Then, it is stored in non-resident storage area 1 (left end in Figure 2).

Then, if the word information that appears belongs to LENGI, its address (index information), if it belongs to LENG 2, the address in I, I!NG 2, h ``FF'', and if it belongs to LENG 3, h ``'FIE''. and LEN
Using the address 4J in G3, the data is stored in the order of small words, resulting in a state as shown in FIG.

As described above, compressed display data and a data structure for restoring and reproducing it are obtained.
rank) is required. For this purpose, grammatical items as shown in FIG. 5 are incorporated into the reproduction algorithm, and the breaks between words are automatically determined.

If we assume that the two consecutive stored contents in the DSP area 32 are a and b, then a, b4 years old, 0'i word, and so on.

There can be 3 types of numbers and symbols, so the silk combination of a and b is 9
As expected. and a for each combination.

Decide in advance whether or not a blank is required between b.
Based on this, the old ANK code is added.

For example, if both a and b are padded, a blank is required (“1”), whereas if the preceding a is a word and b is a symbol, a blank is determined to be unnecessary (“0”). let

Next, the restoration and playback (reproduction) of compressed data will be explained. For convenience of explanation, the following assumptions are made.

Among the elements 0 to 39 of LENGI, the first half elements 0 to 19 are used as a dictionary for numbers, and elements 20 to 39 are used as a dictionary for symbols.

The compressed data has already been transferred to the DSP area and is transferred to Vl? It is assumed that the image is reproduced on the AM area 31.

Let the number of elements in the Dsp area be n, and the current value is −1 (0<k−
It is assumed that elements up to 1<n) are reproduced.

For example, at the CRx address on the VRAM area 31, h “41”
If (“A” in ASCIN expression) is stored, V
I? AM(CRx)=h is written as “41”.

FIG. 6 shows the processing procedure for reproducing only one word from the data compressed and stored in the DSP area 32. Therefore, in order to reproduce all the elements (words) in the r]sp area 32, this processing procedure will be repeatedly executed. Referring to FIG. 6 below, data to be reproduced next is taken out from the DSP area 32 and stored in the variable Cot)E (a). In the next step, the content of the variable cong is compared with hFB'', and if it is greater than hFB, it is determined to be a discrimination code, and the subroutine acce shown in Figure 7 is executed.
Execute ss 2 (mouth).

If it is less than "h FE", it is +, and it is a word, number, or symbol related to gNG1, and the contents of variable Co!1B and 4
0 and discriminate between the two, and if it is less than 40, execute the subroutine access O shown in FIG.
If the altitude is 40m, subroutine a shown in Fig. 7
Execute ccess 1 (c). After executing each buroutine in this way, the grammar processing subroutine 5y
Execute ntax (2).

Next, based on FIG. 7, the subroutine access 0,
1 + 2 will be explained. In Figure 7, the variable SYM
1 indicates whether the code to be reproduced belongs to a word, number, or symbol; in the case of numbers, 0≦SYM I≦19
, in the case of symbols, 20≦SYM 1≦39, and in the case of words, SYM1-40. Note that the variables shown in Figure 8, which will be described later, are SYM for the code reproduced immediately before.
It has the same function as 1.

Variable B indicates a block in the L E N G area, and LPN
In G I, B = O, 1, in ENG2, B = 1, LI
In ENG3, B=2.

The variable A indicates the address of the I'rNT area, and its calculation method is as described in the explanation of procedure H.

The variable Ar1rl indicates the Suf-I address of the rlTcT area.

Sarani r'rNT (A +1), PINT CA
] Lj: 1st position of the DICT area no start address,
It is a combination of lower order parts.

The variable LEN indicates the length of the compressed word.

Now, subroutine access OLl variable SYM I
Stores the value of C0IIE in and returns.

Subroutine access 1 sets variable LEN to 1, EN
Store the value modified by C0DR of G I, that is, the length of the word, set the variable B=O, and then set it to 1ζ as shown in (bo) and (b), and set the variable ADD, that is, the start address of the old CT area. is calculated, 40 is stored in SYM I, and the process returns.

Subroutine access 2 has a determination code h”FE”
.

h FF” is extracted, first find the value of B, then set the next value of the discrimination code to the variable C0
Store the word fragment in DIE (g), and store the word fragment in IJN (ch). Furthermore, in the case of access 1, the start address of the rllcT area is calculated, 40 is stored in SYM 1, and the process returns.

Next, FIG. 8 showing the subroutine 5yntax of the grammar will be explained. The column () shows the processing procedure when the code to be reproduced is a word; in this case, as is clear from Figure 5, regardless of the type of the immediately preceding reproduction code,
It is necessary to create a code for nk. VI there? /I
Increment the address on M area 31), 1
Reserve space for characters.

In addition, at the initial time, that is, the value of nsr + area 32 is set to VR.
When trying to reproduce in the AM area 31, V RAM
It is assumed that all blank codes are stored in the area 31.

6 Next is SY? The value of SVM 1 is stored in lO in preparation for the next cycle (wa).

Next, store the value of the rlTcT area 21 decorated with A(lrl) in the address decorated with flV in CTIX of the VIIAM area 31-hi.And AI)D and CII
Increment X and decrement LEN.

This operation is repeated until IJiN=0 and the process returns.

The (nu) column shows the processing procedure when the code to be reproduced is a symbol, and as is clear from FIG. The value of LENG 1 (symbol code) modified with SYM 1 is stored in the address modified with .

Then increment CRX and set SYM to SYMO.
It stores 1 and returns after preparing for the next cycle.

The column ( ) shows the processing procedure when the code to be reproduced is a number, and in this case, as is clear from Figure 5, when the code reproduced immediately before is a number (SYM O < 20
) only 1) It does not generate rank code, so immediately (
j) Reproduction is performed by processing columns, but if it is a word, CI? After incrementing X to ensure a blank space, start processing the (nu) column. In this way, one cycle of compressed data reproduction is completed.

(6) Effect Next, the effects of the present invention will be quantitatively clarified.

Assume that the display target consists of 300 pages, and the total number of words (counted without duplication, that is, types) is 10.
00, the average word length is 10 characters, and the number of words per page is 100.

(i) When the display target is not compressed It is assumed that one character is expressed in one byte using an ASCII code. Furthermore, if we calculate optimistically without considering the blank code, the storage capacity will be 100 xlox300 =
300,000 (bytes) is required.

(ii) In the case of the method of the present invention, when calculating extremely pessimistically assuming that each word is compressed to 2 bytes, the storage capacity is as follows. However, 40 characters including blank code, numbers, and symbols will be considered.

Compressed data: 100 (words/page) x 2 (hide)
X3000 (page) -60,000 (hyde) 1) ICT area: 10x1000=10.000 (
Hyde) LENG area: 1000+40=1,040
(Hide) PINT area: 2 X100 = 2
,000 (bytes) Then the total is 73,040
Comparing this with 300,000 bytes, the storage capacity can be reduced by at least 75%. The effect of compressing the display data in this way is sufficiently obtained, and since the display is static character information, the time required for reproduction processing does not pose any practical problem, making it extremely practical.

In this way, according to the present invention, by performing data compression, even when displaying a large amount of character information on a display device, a device with a very small walking distance is sufficient, and a system with high storage utilization efficiency is constructed. There are advantages that can be achieved.

[Brief explanation of the drawing]

1. Figure 1 is an external view of a display device to which the method of the present invention is applied, and Figure 2.
The figure is a block diagram showing the circuit configuration, and Figure 3 is a block diagram showing the circuit configuration.
FIG. 4 is a conceptual diagram showing the structure of data stored in the device. FIG. 4 is a conceptual diagram showing the data structure when there are multiple L E N G areas. FIG. 5 is an explanatory diagram of grammar points. 6.7. FIG. 8 is a flowchart showing the processing element 111n for reproduction display. 1-Microprocessor 2-Resident memory (al or 3~work area 4-Non-resident storage area 6-CRT display device 7-Display control circuit patent Applicant Sanyo Electric Co., Ltd. agent Patent attorney Noboru Kono0 LENGI LENG3 Figure 4, page 5

Claims (1)

[Claims] 1. In a data display method for displaying character information stored in an image memory on a display device, words related to the character information to be displayed are stored in a one-dimensional array in a specific area of the storage device, In addition, data relating to the starting address and length of each word in the specific area are stored in another area of the storage device in association with each other, and the character information to be displayed is determined by the string of the word and the length of each word. A data display method characterized by compressing and storing data to be specified as a string in another area, restoring and reproducing required character information based on the stored information, and storing it in an image memory.