JPH0658640B2 - Control method of electronic computer - Google Patents

Description

Description: BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a method of controlling an electronic computer.
Law, especially operating system control table control
Regarding the method. Generally, a user program is stored in a computer.
In, it works in units of work called tasks
Is normal. That is, a task is associated with a program
Resources such as central processing unit and peripheral devices based on the data
It is the smallest unit of work that can be performed. Operate
The loading system gives priority to this task and
Allocate resources more efficiently and manage the task execution state
Each task for efficient scheduling
Management table called TCB (Task Control Block) for each
Is provided. FIG. 1 is a block diagram of this TCB. Main memo
TCB10 in operating system 10 on R1
0 is TCB110 for task number = 1, task number =
TCB115 for 2 and so on, task number = ma
Up to TCB 120 for x, each TCB has the same capacity and
-Mat, lined up in a sequential manner,
Start with fixed address 50 in rating system 10
The address is shown. Here, the task number is the operation
The rating system identifies each task
It is a consistent number assigned to the task in the
It is given when it is created as a task. Next, the details of TCB
The detailed configuration will be described. 200 is a pointer for waiting for various resources
Yes, the resources required by the task are used for other tasks.
While waiting, use this pointer to enqueue
It 205 is a task status (running, waiting for resource availability)
Flags indicating "medium, paused, etc." Among them,
255 and 260 are from main memory and auxiliary memory
System configuration (hereinafter referred to as DISC-equipped system)
This is a flag used in (b). 215 is a book
This is an area for storing the activation factor when the disc is activated. 210
And 220 are computers having a plurality of system configurations
System (hereinafter referred to as multi-computer system)
Link between computers with the information needed in
It is used as an area for taking up the charge. 225 is
Stores task attributes (task type, priority number, etc.)
265 of them are with DISC
This is an area used only for the system. 230
Is information that indicates the position of the task work area.
Relative byte address from the beginning of the store area of the task
The upper and lower limit addresses of the work area are stored in the form of
ing. 235 is the head address in the main memory when this task is operating.
Shows the dress. 240 controls other tasks
This is the information needed to control, 270 of which is D
This information is required only for systems with ISC.
250 is only for multi-computer system
Required information. Each TCB has the above configuration.
The capacity of each is 37 bytes. Here, looking at each information of this TCB
Then, depending on the computer system configuration, unnecessary information may be present.
It 210, 220, 250 are multi-computer systems
This information is required only when the
It is not necessary in the system form. Furthermore, 255, 260, 2
65 and 270 are required only in the system configuration with DISC.
This is important information and is a system configuration consisting of only main memory (below.
It is unnecessary in the main memory only system below)
It That is, the system form is a system with DISC.
System, multi-computer system, main memory only
-Which is different from the system, TCB is
Even in the form, they have the same configuration. The reason for this is explained next
Reveal. FIG. 2 shows one of the TCB 110 shown in FIG.
To extract the data 245 of the register into the register 10 (R10)
The following is an example of a conventional program. In program 300
The instructions used are briefly described. LDR; main memo
Specified by reading one word (32 bits) from the memory
Set in register. LD: Read one word from main memory
Set to Acc register. M: Acc register contents and contents
The product of the contents of the memory is set in the Acc register. TTR;
Sets the contents of the Acc register to the specified register.
AND; Logical product of Acc register contents and memory contents
And the result is set in the Acc register. SR; Acc
Shifts the contents of the register to the right by the specified number of bits. The program 300 requires the following procedures.
is doing. Step 305: Top add of TCB100
Set the address in the R5 register. Steps 310-320; TCB100
37 bytes of TCB for each task number
Case number (1) of TCB 110 including data 245
Et al., Relative address of TCB110 in TCB100
Is set to the IXR1 register. Step 325; 1 containing data 245
Word, ie the 9th word of TCB110 from main memory Acc
Read to register. Steps 330-335; of the extracted Acc
The data 245 is cut out from the content. Step 340: Cut out the data 245
Set in R10 register. Here, the configuration of the TCB is changed for each system form.
And store only the information needed for that system configuration.
Considering the case of the different configuration, as shown in FIG.
The program 300 for fetching byte data 245 is T
The capacity of the CB 110 changes, and the data 245
Because the position of the included 1 word (32 bits) changes,
4 of steps 315, 325, 330, 335
It is necessary to change the place. As described above, according to the conventional technique, the T
If the CB configuration is different for each system type,
You have to change every program that you access
It In other words, a program that accesses the same data
It is necessary to create a system for each system form and avoid it.
Therefore, TCB has the same configuration regardless of the system form.
Was becoming. Therefore, in some form of system
Is the capacity of the TCB, that is, the operation that occupies the main memory.
The problem of unnecessarily increasing the capacity of the operating system
was there. The object of the present invention is to solve the above-mentioned problems.
Therefore, the TCB that actually exists in the main memory differs depending on the system configuration.
It is configured to store only the necessary information.
From the access program, the virtual
Even if it looks like TCB and the system form changes,
Provide a computer control method that eliminates the need to change the RAM
Especially. Means for Solving the Problems A program change is required.
The cause is that the program itself is trying to access
The point is that we are directly aware of the layout address and size of the data.
It The present invention pays attention to this point and adopts the conventional TCB configuration.
Each data is assigned a number, and it corresponds to that number unit
Control data that stores the allocation address and size of the data to be stored.
Bull, and specify only that number in the program
Provide instructions to access the data. [Operation] Even if the TCB configuration is changed for each system form,
A control table that stores addresses and sizes is stored in the system.
Change the program body simply by preparing for each
Unnecessary. Embodiments of the present invention will be described below with reference to FIG.
Reveal. Operating system 10 on main memory 1
The TCB100 has a different configuration depending on the system configuration.
And, the consistent number from 1 is given to the data in TCB, and the
Data allocation address, size, etc. corresponding to the number unit
Stored control table: FDSCR600 and TCB1
00 and FDSCR600 start address, TCB
Control table that stores the capacity of 1 case: TDS
CR500 is provided and TDSCR is made possible by fixed address 400
Indicates the start address of 500. Where FDSCR60
0 is the number assigned to the data (called the field number) =
FDSCR605 of 1, FDS for field number = 2
CR610, FDSCR6 for field number = max
Sequentially in the order of field numbers such as 15
The TDSCR500 is an operating system to which the present invention is applied.
From the number 1 attached to the management table of the operating system
TDSCR5 for consistent number (called table number) = 1
05, TDSCR510 for table number = 2, table
For example, TDSCR515 for number = max
Bulls are sequentially arranged in order. TCB
Bull number is set to 1. The processor 6015 has a register group 6045.
And an arithmetic unit 6030 including an arithmetic unit 6050 and control
Mechanism 6035, control memory 6040, memory interface
Interface mechanism (memory address register: MAR60
20, memory data register: MDR6025)
Composed. The operation of the processing device 6015 is similar to the conventional one.
It is as follows. Next, according to a command from the control mechanism 6035,
The address of the instruction to be executed is via the arithmetic unit 6030
Access request to main memory 1 set in MAR6020
Is issued, the next execution from main memory 1
Command is fetched and set in MDR6025. MD
The instruction set in R6025 immediately returns to the control mechanism 603.
5, decoded there, and stored in the control memory 6040.
Where it is stored, the micro that executes the instruction
Control is transferred to the program. The micro program is
Control memory 6040 sequentially by control mechanism 6035
Read out and decrypted. The control mechanism 6035 has
Based on the calculation mechanism 6030 and the memory interface.
The memory control mechanism 6010. The data operation instruction according to the present invention is a micro
Implemented by programs 700 and 800, control memory 604
Store in 0. FIG. 4 shows details of TDSCR. 550
Is the start address of the management table to be controlled, 56
0 is the data address of the management table to be controlled.
And the size of the FDSCR table that stores the size
Dress 565 is one management table case to be controlled.
Capacity (number of bytes) and 570 are control targets
Indicates the number of cases in the management table. FIG. 5 shows details of the FDSCR. 650
Indicates whether the data is bit data or byte data
Information, 660 is the size of the data (when it is bit data
Is the number of bits, the number of bytes for byte data), 670
From the table top of the data (top of each case)
Relative address (bit address for bit data,
In the case of byte data, it indicates the byte address). Access is made by such a control table.
Data to be managed by numbers, while
Then, the instructions as shown in FIGS. 6 to 10 are provided. FIG. 6 shows a table in which data to be accessed exists.
To know the address of the table; GETER's command
A format 700 is shown. 705 fulfills the above function
Operation code (OPC)
D) and 710 are tables in which data to be accessed exists.
Table number (TBLN), 715 to access
Indicate how many cases the data is in the specified table
Case number (CSEN), 720 is the calculated address
Indicates the register number (REGN) to store. FIG. 7 shows the actual reading of data from the table.
To store or store in table; LDTF / STTF
Shows an instruction format 800 of. 805 is the above function
Operation code indicating that the instruction fulfills
(OPCD), 810 is a table in which the data exists
Table number (TBLN), 815 is read or
Field number (FLDN) of data to be stored, 8
20 is the address (GE) of the table in which the data exists.
Register that stores the address obtained by the TAR instruction)
Number (REGN), and 825 read data
If (LDTF), the register that stores the read data
Data number and data to store (STTF), store
Number of the register that stores the data to be stored (REGN)
Indicates. 8 to 10 show GETAR, LDTF, S
The processing contents of the TTF instruction are shown. FIG. 8A shows a processing flow chart of the LDTF instruction.
8B, the data in the LDTF instruction
Indicates the operation content. Below, in FIG. 8 (a) and FIG. 8 (b),
Therefore, the processing flow of the LDTF instruction is changed to that of the HARD machine of FIG.
It will be explained in relation to the structure. Other instructions; STTF, GETAR
However, since it can be inferred from the explanation of this LDTF instruction,
The process flow will be explained only. A (xxxxx) in FIG. 8B is xx
Indicates an address of ×. Block 1005; by control mechanism 6035
And fetched instruction TBLN810 and memory control mechanism
Top add of TDSCR500 obtained through 6010
Res is calculated by the arithmetic unit 6030 and obtained
Address 55 of TDSCR to RE of register group 6045
Store in Gi. 51, 52, 53, 54, 55 Block 1010; Control mechanism 6035 uses register group 6
045 REGi contents, that is, the corresponding TDSCR address
Accessing the dress via the memory control mechanism 6010
FDS by TPFDS560 of TDSCR500
Find the start address of CR600. Block 101
5; Command field fetched by the control mechanism 6035
De FLDN815 and the previously obtained FDSCR600
The start address is calculated by the arithmetic unit 6030 and
Get the address 59 of the FDSCR. Obtained address 5
9 is stored in REGj of the register group 6045. 5
6, 57, 58, 59 block 1020; control mechanism 6035 uses register group 6
The contents of REGj of 045, that is, the address of the corresponding FDSCR
The dress 59 is accessed via the memory control mechanism 6010.
To the DATATYPE 650 in the FDSCR600.
Therefore, the field data is byte data.
Or bit data is determined. Block 1025; the field data B
If 150 is byte data,
The instruction field REGN820 extracted from
Main memory address of this table body 100 and REGj
Performed by DATAADDR670 in the applicable FDSCR
Applicable field data added by arithmetic unit 6030
Obtain the main memory address 61 of B150. Obtained ad
The address 61 is stored in REGk in the register group 6045.
It 60, 61 Block 1030, the field indicated by REGk
After the address in the main memory of data B150, FDSCR
Data for DATASIZE660 in 600
Is accessed via the memory control mechanism 6010,
The field data B150 is obtained. 62, 63 block 1035; the obtained data B63 is used as a controller
Command field REGD8 fetched by structure 6035
25, and the process ends. (REGD is a register group
1 register 6045) 64, 65 block 1040; the field data A160 is a bit
In the case of automatic data, first, REG of the register group 6045
DATAA out of the address 59 of the FDSCR indicated by i
The contents of the DDR unit 670 are stored in the memory access mechanism 6010.
Obtained via and multiplied by 1/8 by the arithmetic unit 6030
The byte address of the field data A160
For the instruction fee fetched by the control mechanism 6035.
The main of the relevant table body 100 indicated by the REGN 820
The memory address is added by the arithmetic unit 6030.
And the main memory address of the corresponding field data A160.
You get the 61. The obtained address 61 is the register group 6
Store in REGk in 045. 60, 61 block 1045; the field indicated by REGk
4 bytes after address 61 in main memory of data A160
Minute data is accessed via the memory control mechanism 6010.
4 bytes including the corresponding field data A160
Data 67 is obtained. 66, 67 Block 1050; Add of the FDSCR indicated by REGj
Make a note of the contents of DATA ADDR section 670
Obtained via the re-access mechanism 6010, the arithmetic mechanism 603
The field data obtained by multiplying by 1/8 by 0
Data A160 bit address (1/8 times the remainder is
Address) and DATASI in FDSCR
According to ZE660 (bit size for bit data)
The 4-byte data 67 including the data A previously obtained
Of the data, only the data A160 is calculated by the arithmetic unit 6030.
Cut out. 68 block 1055; computing the obtained data A160
Shift right by the structure 6030, and (right right) control mechanism 60
35 into the instruction field REGD825 fetched by
Store and end the process. (REGD is register group 6
1 register 045) 69, 70 FIG. 9 is a processing flow chart of the GETAR instruction. First specify
The start address of the TDSCR corresponding to the specified TBLN
Ask (step 905). In the TDSCR
Is the head of the table from CSEN specified as SIZE
Calculate the relative address from the specified case to the specified case (S
Calculated by IZEX (CASE-1)) (Step 91
0). Next, the relative address obtained above and the TDSCR
Table head address of
Obtain the address (step 915). Designated by the above
Specify and specify the absolute address of the specified case in the table
Stored in the registered register (step 920)
Processing is completed. FIG. 10 is a processing flow chart of the STTE instruction.
It 1105 to 1120 are the same procedures as the LDTF instruction
Determines whether the data is bit data or byte data
It If it is byte data, the address of the specified table
Data (content of REGN) and DATAAD in FDSCR
Add DR and obtain the byte address of the data 112
5. DAT in FDSCR with information specified by REGD
Find the number of bytes specified by ASIZE above
The data is stored after the address of the data 1130, and the processing is performed.
finish. For bit data, DA in FDSCR
From TAADDR (bit display), specify the byte address
Calculate and add the address indicated by REGN
1135 for the byte address of the. The value obtained above
4 bytes of information after the right address
Read to querier. DATA ADDR in FDSCR and
Obtain the bit position of the data by DATASIZE
Therefore, the data portion is erased from the read information 114
5. Delete the write data specified by REGD above.
1150 to embed in the left part. The obtained 4-byte data
Re-store the data after the byte address when the data was read 11
55, the processing is ended. The data numbers to be accessed are
Control table that stores data, address and size of data
And set the instruction to access the data by specifying the number.
1 byte of the data of FIG. 1 shown in FIG.
A conventional program 300 for reading data is shown in FIG.
Program 6000. Here, the byte data in Figure 1
Even if the position and size of the data changes,
DATASIZE and DATAAD in FDSCR
6000 programs in Figure 11 just by changing the DR
You can access the same data without changing your body.
And are possible. Now, by making the above basic motives,
The TCB has the following configuration. FIG. 12 is a view from the accessing program.
2 shows the configuration of the TCB 1500 according to the present invention. Shown in Figure 1
Data 1 in the TCB in the conventional TCB configuration 110
Add fields 1 to 44 to each one. This
As described in FIG. 1, 255 (field number =
8, 12 to 17) and 260 (field number = 2)
0) and 265 (field number = 24 to 30,3)
1, 33, 34) and 270 (field number = 4)
0) is the information required only in the system configuration with DISC
And 210 (field number = 18) and 220
(Field number = 22) and 250 (Field number
No. = 44) in the multi-computer system form
This information is necessary only for you. A program that accesses like this
The TCB seen from the ram is the same regardless of the system configuration.
I will keep it as one. This TCB1500 does not actually exist in main memory.
It does not exist and is called a virtual TCB (VTCB). FIGS. 13 to 15 show TCs which actually exist in the main memory.
The structure of B (this TCB is called RTCB) is shown. FIG. 13 shows R in the system configuration with DISC.
The TCB configuration is shown. From VTCB in Fig. 12, multicon
Information required only for computer system form (field number
No. = 18,22,44)
The configuration. FIG. 14 shows a multi-computer system form.
The RTCB structure at this time is shown. This RTCB is VTCB.
Same configuration as. FIG. 15 shows the main memory only system configuration.
Shows RTCB configuration 3000, DI from VTCB
System form with SC and multi-computer system
Information necessary only for form (field number = 8, 12 ~
17, 20, 24-30, 31, 33, 34, 40, 1
(8,22,44) is removed and one case is composed of 26 bytes
To do. FDSCR is created for each RTCB.
It FIG. 16 shows F in the system configuration with DISC.
Indicates a DSCR. For field number (FLDN) = 1
FDSCR3005, = 2 for FDSCR3010
To make FDSCR3050 for 44
However, the contents of each FDSCR are RTCB ・ TYPE1 (Fig.
13) is created. For example, FLDN = 42
FDSCR3040 has a data type (DATATYPE
E) is byte data, size (DATASIZE) is 1
Byte, address (DATAADDR) is 29 bytes
Create it like an eye. There is on VTCB,
Data not found in RTCB / TYPE1 (FLDN = 1
8,22,44) FLDN3020,303
0, 3050 sets the content as “0” and sets the FLDN
It is clearly stated that there is no data for this RTCB. FIG. 17 shows a multi-computer system form.
8 shows an hourly FDSCR 3100. RTCB ・ TYPE
FLDN = 1 for 3105 to FL as in 1 (FIG. 16)
Created up to 3120 for DN = 44, and the contents of each are R
Created to represent TCB / TYPE2 (Fig. 14)
It FIG. 18 is for the main memory only system mode
FDSCR3200 of FIG. FLDN = same as the former
Created from 3205 for 1 to 3325 for FLDN = 44
The contents of each are shown in RTCB ・ TYPE3 (Fig. 15).
Create to forget. For example, FDS for FLDN = 42
CR3320 is DATATYPE and DATASIZE
Is the same as the former, but DATA ADDR is 23
It becomes the eye. It is on VTCB, this RTCB / TY
Data not found in PE3 (FLDN = 8, 12-18, 2
0, 22, 24 to 31, 33, 34, 40, 44)
FDSCR3215, 3220-3250, 325
5,3260,3265-3300,3305,331
0, 3315, and 3325 have the content set to “0”,
There is no data for this RTCB in this RTCB.
Be explicit. FIG. 19 shows the FDSCR for each system form.
The relationship of RTCB is shown. In TDSCR505 for TCB
TCB start address (TPTCB) 4000 of
400 or RTCB / TYPE1 depending on the system type
5, pointing to RTCB / TYPE2 or 4010, or R
TCB / TYPE3 or 4015
Select when narrating. Also in TDSCR505
The TPFDS5000 of 500 is also 500 depending on the system configuration.
5 or 5010 or 5015 system generation
Select at times. SIZE565 in TDSCR505
Is a value of 32 when RTCB / TYPE1 is selected
If you select TYPE2, set the value of 37 to TYPE.
When 3 is selected, the value of 26 is also set.
Embed at the time of translation. As described above, the FDSCR for each system type
Create the system form during system generation
Select and access the FDSCR and RTCB that match
The program should be aware of only the VTCB structure.
Changes the TCB configuration without changing the program.
It is possible to change. In this embodiment, the minimum RTCB is used.
In the case of the main memory only system, the TCB capacity is
11 bytes per task compared to the TCB capacity of
You can do it. The effects of the present invention are shown below. 1) Operating system for managing tasks
The effect of the TCB configuration in the system on the access program
It is possible to change without. 2) Accordingly, depending on the system configuration,
Unnecessarily increasing the capacity of TCB, and eventually the main memory
Control the increase in operating system capacity
Can increase the main memory range available to the user
It has become possible. (Maximum of 1
You can reduce 11 bytes per disk,
2805 bytes capacity when the maximum number of packs is 255
Can be reduced)

BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is a conventional TCB configuration diagram. FIG. 2 is a diagram showing an example of a conventional TCB data access program. FIG. 3 is an overall view of a table provided according to the present invention. FIG. 4 is a diagram showing details of FIG. FIG. 5 is a diagram showing details of FIG. FIG. 6 is a diagram showing details of FIG. 3; FIG. 7 is a diagram showing details of FIG. 3; FIG. 8 is a processing flow chart of instructions. FIG. 9 is a flow chart of processing an instruction. FIG. 10 is a flow chart of processing an instruction. FIG. 11 is a diagram showing an example of a TCB data access program according to the present invention. FIG. 12 is a configuration diagram of a TCB modified according to the present invention. FIG. 13 is a configuration diagram of a TCB modified according to the present invention. FIG. 14 is a configuration diagram of a TCB modified according to the present invention. FIG. 15 is a configuration diagram of a TCB modified according to the present invention. FIG. 16 is an example diagram of a table provided in connection with the present invention. FIG. 17 is an example diagram of a table provided in connection with the present invention. FIG. 18 is an example diagram of a table provided according to the present invention. FIG. 19 is an example diagram of a table provided according to the present invention. [Explanation of Codes] 500 ... TDSCR, 600 ... FDSCR, 700 ... GETAR instruction, 800 ... LDTF / STTF instruction, 1500 ... Virtual TCB, 2000 ... RTCB / TYPE1, 2100 ... RTCB / TYPE2, 2200 ... RTCB / TYPE3, 3000 ... FDSCR / TYPE1, 3100 ... FDSCR / TYPE2, 3200 ... FDSCR / TYPE3.

Claims (1)

[Claims] 1. In a computer control method for storing a plurality of tasks, which are units of work generated from a user program, in a main memory and operating the tasks on an arithmetic processing unit, a plurality of tasks necessary for controlling each of the tasks are stored. A plurality of task management tables, each of which stores data in a plurality of data fields and has a data field configuration according to the type of the plurality of stored data, are provided in the main memory for each of the plurality of tasks. Storing the identification information of each of the plurality of data as data field information of a virtual data field configuration defined in advance separately from the data field configuration, and for each of the data field information, the relevant data Stores the first allocation information of the data field in each tax management table that stores For each of the first control table and each of the plurality of task management tables, a second control table storing second allocation information of the task management table in the main memory is stored in the main memory, and the calculation is performed. The processing device designates one of the plurality of task management tables and uses the second control table to obtain the second placement information of the designated task management table, and By specifying the specific data field information and using the first control table, the first arrangement information relating to the specific data field information is obtained, and further, by using the second arrangement information, A control method for an electronic computer, characterized by executing a second instruction for reading and writing the corresponding data. 2. The first layout information is a data format, a layout address, and a size of the data field, and the second instruction has the layout address according to the data format,
The method of controlling an electronic computer according to claim 1, wherein the data of the size is read. 3. 2. The computer according to claim 1, wherein the second allocation information is, for each of the plurality of task management tables, a start address, a table size of one case, a total number of cases, and a start address of the first control table. Control method. 4. The first address of the second control table is stored in the fixed address area of the main memory, and the first address is stored.
4. The method of controlling an electronic computer according to claim 3, wherein in the instruction, the start address of the fixed address area is further used to obtain the start address of the corresponding case of the specified tax management table. 5. A plurality of the first control tables are stored for each of a plurality of types of the data field configurations, and the arithmetic processing unit sets one of the plurality of data field configurations according to the system configuration at the time of system generation. 2. The method of controlling an electronic computer according to claim 1, wherein the content of the second control table is set according to the selected data field configuration.