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C instruction time

 
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jksor1234



Joined: 18 Jan 2007
Posts: 9

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C instruction time
PostPosted: Thu Jan 18, 2007 9:06 am     Reply with quote

Is there a document which outlines how much time it takes to execute every 'C' instruction.

In the datasheet for the PIC16F913, it states that the Instruction Cycle Time equals four divided by the oscillator frequency --> Tcy = 4/Fosc

I want to know how many instructions an IF statement uses, how many a WHILE loop uses, how many an Interrupt Routine uses, etc. In the CCS Manual, it gives a table for the amount of time math functions take.

Is there a document which says how many instruction cycles each instruction takes or do I need to just go through the LST file and figure it out?
Ttelmah
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PostPosted: Thu Jan 18, 2007 9:25 am     Reply with quote

Simple answer. No.
It will even vary according to the data involved, and where the instructions occur.
There is a table in the manual, giving 'typical' times for the maths operations, but the problem is that a single C instruction, may involve anything from one assembler instruction to thousands. The compiler optimiser, is smart enough so that (for instance), if a page switch is necessary in one location, but not in another, it is removed in the latter case, so exactly the same access instruction, can have different numbers of machine instructions in different locations. As another example, taking the 'modulus' (remainder after division), of two numbers. This according to the timing 'table', takes typically 116 instructions for two 8 bit numbers. However the latter optimsers are smart enough, that if the second number is a 'binary' value (2, 4, 8 etc.), and a constant, the '&' operand is used instead, taking just a couple of instructions.
The only way to find out, is to either manually count instructions, or use a tool like the MPLAB simulator, and sopwatch function, and physically time the operation for real. Even here you need to be aware that timings for arithmetic operations in particular, will change with different values...

Best Wishes
rwyoung



Joined: 12 Nov 2003
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PostPosted: Thu Jan 18, 2007 9:27 am     Reply with quote

Use the LST file.

The time to complete an if or while will vary with the argument list. Is your argument list a simple variable comparison (i<3) or a more complex multi-level with pointers? Also things like chip archetecture can affect the time.

This has been suggested MANY times and I'll say it again. Learn a little assembly first. You don't have to be proficient, but at least be able to read it and recognize the basics. Then when you are working with the CCS compiler and you just KNOW your C code is correct but yet the wrong thing happens, you have a chance. Look at the LST file, see what the compiler produced. It does occasionally generate "crap" (usually though it is a mater of GIGO).
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Rob Young
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jksor1234



Joined: 18 Jan 2007
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PostPosted: Thu Feb 08, 2007 4:05 pm     Reply with quote

Where in the LST file can you see the instructions required to get into an interrupt and the instructions required to get out of an interrupt? I want to calculate the amount of time my interrupt routines are taking.
Ttelmah
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PostPosted: Thu Feb 08, 2007 4:19 pm     Reply with quote

Comment out the line '#nolist' in the include file for the processor you are using. This allows the 'hidden' bits of housekeeping code to be seen.
Then just look at the code starting at the hardware interrupt address (depends on whether this is a PIC16, or a PIC18). Expect perhaps 30 to 40 instruction times, before the actual jump to the handler code, and a similar number on the return.

Best Wishes
jksor1234



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PostPosted: Fri Feb 09, 2007 8:55 am     Reply with quote

I am using a PIC16F913. Where is the hardware interrupt address?

From the LST file:
Code:
.................... #int_TIMER1
.................... void TIMER1_isr(void)       //setup to interrupt every 262mS
.................... {
....................    Check_Flag = TRUE;      
*
0073:  BSF    2A.0
.................... }
.................... 
.................... 
0074:  BCF    0C.0
0075:  BCF    0A.3
0076:  GOTO   022

Is 0022 the interrupt address?

at 0022 in the LST file:
Code:
0022:  MOVF   22,W
0023:  MOVWF  04
0024:  MOVF   23,W
0025:  MOVWF  77
0026:  MOVF   24,W
0027:  MOVWF  78
0028:  MOVF   25,W
0029:  MOVWF  79
002A:  MOVF   26,W
002B:  MOVWF  7A
002C:  MOVF   27,W
002D:  MOVWF  7B
002E:  MOVF   28,W
002F:  MOVWF  0A
0030:  SWAPF  21,W
0031:  MOVWF  03
0032:  SWAPF  7F,F
0033:  SWAPF  7F,W
0034:  RETFIE
0035:  BCF    0A.3
0036:  GOTO   073


So, when the interrupt occurs, the PIC jumps to 0022. It sees if the Timer1 caused the interrupt. If it did, then it jumps to 0073 and runs the Timer1 interrupt routine. After the routine is completed, it jumps back to 0022 and runs that code again until it sees RETFIE. So, to get into the interrupt routine, it takes 21 instructions. It takes 4 instructions to run the interrupt routine and return to the interrupt handler. And finally, it takes 19 instructions to exit the interrupt handler.

So the total time the interrupt will take, using an 8MHz processor is...
21+4+19 = 44 instructions
500nS per instruction
22uS total execution time.

If more than one interrupt is enabled, then the number of instructions to get in and out of the interrupt routine will change because now the interrupt handler has to figure out which interrupt occured.

Are the above statements correct?
Ttelmah
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PostPosted: Fri Feb 09, 2007 9:51 am     Reply with quote

The hardware address for a 16 chip, is 0004.
Look at the data sheet...
You will find elsewhere in the assembly, the 'global' handler. This saves all the registers, then jumps to the individual hanlder (in your case at 0073). Then the code at here, jumps back when it finishes, which restores the registers, and returns to the caller.
The typical 'global' handler, for a 16 chip, is:
Code:

0004:  MOVWF  7F
0005:  SWAPF  03,W
0006:  CLRF   03
0007:  MOVWF  21
0008:  MOVF   0A,W
0009:  MOVWF  20
000A:  CLRF   0A
000B:  MOVF   04,W
000C:  MOVWF  22
000D:  MOVF   77,W
000E:  MOVWF  23
000F:  MOVF   78,W
0010:  MOVWF  24
0011:  MOVF   79,W
0012:  MOVWF  25
0013:  MOVF   7A,W
0014:  MOVWF  26
0015:  MOVF   7B,W
0016:  MOVWF  27
0017:  BCF    03.7
0018:  BCF    03.5
0019:  MOVLW  8C
001A:  MOVWF  04
//Up to here, is the 'saving' code, and code to set the banks
//for the next part - total of 23 instructions

001B:  BTFSS  00.3
001C:  GOTO   01F
001D:  BTFSC  0C.3
001E:  GOTO   042
001F:  MOVLW  8C  //This part handles the test for an interrupt,
//and jumps to it's handler. In this case another 4 instructions

//From here is the 'restore' code - this is the part you have found
0020:  MOVWF  04
0021:  BTFSS  00.1
0022:  GOTO   025
0023:  BTFSC  0C.1
0024:  GOTO   03F
0025:  BTFSS  0B.4
0026:  GOTO   029
0027:  BTFSC  0B.1
0028:  GOTO   03C
0029:  MOVF   22,W
002A:  MOVWF  04
002B:  MOVF   23,W
002C:  MOVWF  77
002D:  MOVF   24,W
002E:  MOVWF  78
002F:  MOVF   25,W
0030:  MOVWF  79
0031:  MOVF   26,W
0032:  MOVWF  7A
0033:  MOVF   27,W
0034:  MOVWF  7B
0035:  MOVF   20,W
0036:  MOVWF  0A
0037:  SWAPF  21,W
0038:  MOVWF  03
0039:  SWAPF  7F,F
003A:  SWAPF  7F,W
003B:  RETFIE
//Another 28 instructions to return to the caller...


Best Wishes
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