JPH054043Y2 - - Google Patents

Description

[Detailed explanation of the idea]

[Technical field of invention] This invention relates to a program computer. [Prior Art] Conventionally, when a program calculator runs a program that causes a mathematical error such as PO:√ (1-2) (where PO indicates the program number),
When the user notices that an error has occurred and tries to display the error location by operating the cursor keys, a non-error location ")" will be displayed blinking instead of the original error location "√". I was getting used to it. This is to indicate the error location by the contents of the current execution pointer, that is, the 1 in parentheses.
This is because the -2 operation is performed first and then the √ calculation is performed, so the execution pointer is at the closing parenthesis ")". [Problems with the Prior Art] Therefore, since the actual error location is not displayed, there is a problem in that it takes extra time to search for the error location in the case of a complicated calculation formula. [Purpose of the invention] The object of this invention is to provide a program computer that can indicate the operator at the location of an error that has occurred during program execution, and can quickly unpack the program. [Key Points of the Invention] The key point of this invention is to stack operators, pair the address of the operator in the program with the operator, and stack the operators, so that if an error occurs in a calculation using the stacked operators, Specifically, the operator in the calculation formula is displayed in a special manner based on the address stored in pair with the operator. [Example] An example will be described below with reference to the drawings. In FIG. 1, the key input section 1 includes a numeric keypad 1A,
Decimal point key 1B, equal key 1C, +, -, ×, ÷
Four arithmetic operation keys 1D, clear key 1E, all clear key 1F, right parenthesis key 1G, left parenthesis key 1
H, √ key 1I, error release key, calculation execution
RUN keys 1K1J, etc. are provided. The output of each key is then input to the control unit 2 and processed. This control section 2 stores a control program for controlling the operation of the entire computer, and gives control instructions to each section within the circuit. Note that the result data DATA obtained when the calculation section 6 executes the difference calculation is sent to the control section 2 and used for reading the next control command. Numerical data input from the key input unit 1 is written to the X register 3 of the X register 3, Y register 4, and Z register 5 via the control unit 2, and is further sent to the calculation unit 6 where it is used for calculations. Ru. If an error occurs in the calculation, the calculation section 6 sends an error signal to the control section 2. Also, the display data is set
The display data in the register 3 is sent to the display section 8 via the display control section 7 and displayed. 9 in the figure is a program RAM (random access memory), which is addressed by a program counter 10 that is incremented by the control unit 2, and program data for formula calculation is written by key operations on the key input unit 1. be included.
This program data is then sent to the arithmetic unit 6 and executed, but at the time of execution, the operator in the program and the address of that operator in the program are paired and stored in the function stack (FS). 11 and pointer stack (PS) 12, respectively. Further, if an error occurs during execution of the above program, the address of the operator that caused the error is set in the error pointer (EP) 13. When the error release key 1J is operated, the program data in the program RAM 9 is transferred to the display section 8 and displayed, and the cursor is placed under the operator that caused the error on the display section 8 based on the address in the error pointer 13. is displayed blinking. The function stack 11 and pointer stack 12, which are paired together, are connected to the control unit 2.
It is addressed by a stack count (SC) 14 which is incremented (plus, minus) by one. Next, the operation will be explained with reference to FIGS. 2 to 6. First, program data is written into the RAM 9 by operating the keys on the key input section 1. in this case,
The RAM 9 is addressed by a program counter 10 which is incremented by the control section 2, and the program data is written therein. During this writing, this program data is also sent to the display section 8 and displayed. Assume now that a program √ (1-2) containing an error as shown in FIG. 4 is written as a program PO with program number "0". When the RAN key 1K is operated, the processing shown in the flowcharts of FIGS. 2A and 2B starts and calculations are executed. In addition,
As shown in Figure 4, the above program runs on RAM9.
It is assumed that the data is written to addresses 0 to 5 of . When the calculation is started, first, SC14 and RC10 are cleared by the process of step _A1 . Next, in step A, data “√” at address 0 of RAM9
is transferred to Z register 5. Then, it is determined whether the content is a numeric value, an operator, or no code (no data) (step A ₃ ), and since it is an operator in this case, the process proceeds to step A ₅ , in order of priority of operations (the numbers in parentheses are ) is determined. In this case, it is not a priority, so proceed to step A ₆ and 0 in FS11.
The operator "√" is set at the address (states 1 and 2 in FIG. 6). Next, in step _A7 , the data on PC10 is set to “0”.
is set to address 0 of PS12, and then SC14 is incremented by +1 in step _A8 to become "01".
Further, in step _A13 , PC10 is also incremented by +1 to become "1", and the process returns to step _A2 . Next, data "(" at address 1 of RAM 9 is set in Z register 5, and the process proceeds to step _A6 via steps _A3 and _A5 , resulting in the state shown in FIG. 6.3.Next, steps _A7 and _A8 , _A13 sets data "01" in S12, "02" in SC14, and "02" in PC10, and returns to step A2.Then, data " ₁ " at address 2 of RAM9 is set in Z register 5. , the process proceeds from step A ₃ to step A ₄ where arithmetic processing is executed in the arithmetic unit 6, proceeds to step A ₁₃ , where "03" is written to the PC 10, and the process returns to step A _2. Then, the Z register 5 is
Data “-” at address 3 of RAM9 is set,
Proceeding from steps _A3 and _A5 to _A6 , data "-" is set at address 2 of FS11, and step _A7 ,
Data “03” is placed at address 2 of PS12 by A ₈ and A ₁₃ .
However, "03" is set to SC14, "04" is set to PC10, and the state shown in Fig. 6-4 is reached.
Return to _A2 . Next, in step _A2 , RAM9 is stored in Z register 5.
The data “2” at address 4 is set, and steps A ₃ , A ₄ , and A ₁₃ are executed, and the PC 10 becomes “05”.
Then, return to step _A2 . Next, in step _A2 , data ")" at address 5 of RAM9 is set in Z register 5, and step
Data ")" is placed at address 3 of FS11 by A ₃ , A ₅ , and A ₆
is set. Further, in steps _A7 , _A8 , and _A13 , data "05" is set in address 3 of the PS 12, "04" in the SC14, and "06" in the PC 10, respectively, and the process returns to step _A2 . Next, in step _A2 , data "0" (no code) is set in the Z register 5, so after step _A3 , the process advances to step _A15 , and SC14 is -
1 and becomes "03", then step A ₁₆ becomes Z
Data ")" at address 3 of FS11 is set in register 5, and the process advances to step _A18 via step _A17 , where data "04" at address 3 of PS12 is set in EP13. Then proceed to step A ₁₉ , but since there is only ")", no calculation is performed, and step A ₁₅
Return to Then, SC14 is incremented by 1 to become "02", and in step _A16 , data "-" is set in the Z register. Therefore, in steps A ₁₇ , A ₁₈ , and A ₁₉
Data “03” is set in EP13, and the step
Return to A ₁₅ and SC14 becomes "01". Next, in step _A16 , the data "(" is set in the Z register 5, and the operation "1-2" is executed by executing steps _A17 , _A18 , and _A19 , and the resultant data "-1" is stored in the X register. It is set to 3. Also, “01” is set to EP13, and the step
Returns to A ₁₅ , and SC13 becomes "00". Next, in step _A16 , data "√" is set in the Z register 5, and when the process proceeds to step _A18 via step _A17 , data "00" is set in EP13, and then in step _A19 , the data "√" is set in the Z register 5. Since the calculation is executed, an error occurs and the process proceeds to step _A21 , where an error display (for example, "E") is displayed on the display section 8, and the process ends. Then, if you operate the error release key 1J, the third
The process shown in the flowchart of the figure starts, and in step _B1 , the mathematical formula data in addresses 0 to 5 of the RAM 9 are successively output and sent to the display section 8, where they are displayed. Next, based on the data “00” of EP13,
The cursor blinks under the 0th digit data "√" as shown in FIG. 5, and the operator that caused the error is displayed. I have not explained steps A ₉ to A ₁₂ and A ₂₀ in Figures 2A and B, but this is a calculation formula that does not cause an error. Executed when operators such as high x are stacked. Since it is not directly related to the gist of this invention, the operation will be briefly described below with changes in the data set in the Z register 5 and the contents of SC14 and PC10.

[Effect of idea]

As explained above, this idea stacks operators when executing a program, pairs the address of the operator in the program with the operator, and stacks the operators, causing errors in calculations using the stacked operators. If a program error occurs, the program computer is designed to display the operator in a special manner based on the address stored with the operator, so it is easy to identify the operator that caused the error when the program was executed, and to correct the program. There is an advantage that depacking can be done quickly.

[Brief explanation of the drawing]

FIG. 1 is a circuit diagram of a program computer according to an embodiment of this invention, FIGS. 2A and B, and 3 are flowcharts, and FIG. 4 is a diagram showing an example of a mathematical formula written in the program RAM 9. FIG. 5 is a diagram showing an example of a display when an error occurs, and FIG. 6 is a diagram showing changes in the status of each part. 1...Key input section, 1J...Error release key,
1K...RAN key, 2...Control unit, 3, 4, 5
...Register, 8...Display section, 9...Program
RAM, 10...Program counter, 11...
Function stack, 12... pointer stack, 14... stack counter.

Claims (1)

[Claims for Utility Model Registration] It has a program memory and a program counter that sequentially addresses the program memory, and reads key data representing a calculation formula stored in the program memory sequentially according to the program counter, In a program calculator that performs calculations, there is a stack memory that stores operators and the addresses of the operators, and if the key data read above is an operator, the operator and the operator are stored after determining the priority order. a writing means for storing an address in the program memory in the stack memory; and if an error occurs in calculation using the operators stored in the stack memory, displaying the calculation formula and displaying the corresponding address stored in the stack memory; A program calculator comprising: error display means for displaying an operator in which an error has occurred based on the address of the operator in a special manner.