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anestho
Joined: 27 Dec 2006
Posts: 28
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| Fast multiply without hardware multiplier |
 Posted: Sun Sep 09, 2007 1:45 pm |
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Hi,
I was reading the thread below:
http://www.ccsinfo.com/forum/viewtopic.php?t=6131&highlight=fast+multiplication
I know there is a function _mul (x,y) where x and y is 8 bit and the result is 16 bit. I am confused on how this is done on 12F chips with 14bit core without the assembly MULWF or MULLW instruction.
I looked at the .lst file, but it calls the mult8*8 function. Not sure if it stores the high byte of the result. Could someone confirm this? Does _mult work with 12F chips????
Should I just stick to 16 bit result = 16 bit * 16 bit ?
Will 16 bit var = _mul(8bit * 8bit) be equal to above if both the 16bit operand and the 8bit operand are equal?
How can I stick and debug the file in asm? I can do it with CC5x since it generates asm, but CCS doesn't.
I am looking for the fastest possible multiplication of 8*8 bit. |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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anestho
Joined: 27 Dec 2006
Posts: 28
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| Posted: Sun Sep 09, 2007 5:54 pm |
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Thanks, PCM _programmer;
Here is what I found
| Code: |
;***************************************************************************
;** time efficient multiplication 8 bit x 8 bit = 16 bit (unsigned)
;**
;** company: elektronik 21 GmbH
;** programmer: Martin Schaefer (idea from Andrew Warren)
;**
;** execution time: fixed 38 cycles (with jump in and jump out) !!!
;** code length: 35 words
;** multiplier: w
;** multiplicand: resultlo
;** result: resulthi:resultlo
;***************************************************************************
MUL8X8 CODE
Mul8x8 ;* 2 cycles for call - instruction
GLOBAL Mul8x8, resulthi, resultlo
mult MACRO
btfsc STATUS,C
addwf resulthi,F
rrf resulthi,F
rrf resultlo,F
ENDM
clrf resulthi ;* 1 cycle
rrf resultlo,F ;* 1 cycle
mult ;* 4 cycles
mult ;* 4 cycles
mult ;* 4 cycles
mult ;* 4 cycles
mult ;* 4 cycles
mult ;* 4 cycles
mult ;* 4 cycles
mult ;* 4 cycles
retlw 0 ;* 2 cycles
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I am having trouble converting it to CCS #asm format.
I am using a PIC12F683
I defined these at the beginning of the code.
#byte STATUS = 0x03
#bit CARRY = STATUS.0
#bit ZERO = STATUS.2
Here is what I got so far:
| Code: |
/// *************************************************************
//multiplier = W (accumilator)
//multiplicand= result_lo
//result: result_hi:result_lo
int16 mult8x8 (int8 result_lo, char W)
{
int8 result_hi=0;
int16 result = 0;
#asm
rrf result_lo,F
// #1
btfsc STATUS,CARRY
addwf result_hi,F
rrf result_hi,F
rrf result_lo,F
// #2
btfsc STATUS,CARRY
addwf result_hi,F
rrf result_hi,F
rrf result_lo,F
// #3
btfsc STATUS,CARRY
addwf result_hi,F
rrf result_hi,F
rrf result_lo,F
// #4
btfsc STATUS,CARRY
addwf result_hi,F
rrf result_hi,F
rrf result_lo,F
// #5
btfsc STATUS,CARRY
addwf result_hi,F
rrf result_hi,F
rrf result_lo,F
// #6
btfsc STATUS,CARRY
addwf result_hi,F
rrf result_hi,F
rrf result_lo,F
// #7
btfsc STATUS,CARRY
addwf result_hi,F
rrf result_hi,F
rrf result_lo,F
// #8
btfsc STATUS,CARRY
addwf result_hi,F
rrf result_hi,F
rrf result_lo,F
#endasm
result = result_hi*256+result_low;
return result;
}
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I get an error on the 2nd line of the #asm addwf result_hi,F saying "expression must evaluate to a constant"
I am using version 3.249 PCWH compiler. |
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anestho
Joined: 27 Dec 2006
Posts: 28
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| Posted: Sun Sep 09, 2007 6:01 pm |
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This does the same thing as:
| Code: | uns16 operator* _multM8_8( uns8 arg1, char W)
{
uns16 rval = 0;
Carry = 0;
#define addRR(rval,arg,bit) if(arg&((uns8)1<<bit)) rval.high8+=W; rval=rr(rval);
addRR( rval, arg1, 0);
addRR( rval, arg1, 1);
addRR( rval, arg1, 2);
addRR( rval, arg1, 3);
addRR( rval, arg1, 4);
addRR( rval, arg1, 5);
addRR( rval, arg1, 6);
addRR( rval, arg1, 7);
#undef addRR
return rval;
} |
[/code]
this is from CC5x math lib MATH16M.h
If you can get either to work, that would be great, or at least some suggestions.
Here is how I understand it:
1. compare the 0 bit with 1, if it is 1, then add 1 to the MSB and rotate right the bits
2. same thing with bit 1.. etc to bit 7 |
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anestho
Joined: 27 Dec 2006
Posts: 28
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| Posted: Sun Sep 09, 2007 6:53 pm |
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I got it to compile finally. Mistake was using STATUS, CARRY instead of STATUS,0.
Will report if it works. |
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anestho
Joined: 27 Dec 2006
Posts: 28
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| Posted: Sun Sep 09, 2007 7:02 pm |
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Here is what I am using now. Still need to test it, but it compiles.
| Code: |
// Begin Multiplier Routine
int16 mult8x8(int8 mulcnd, int8 mulplr) // 35 instruction cycles = 0.5 x 35 = 17us
{
int8 H_byte = 0;
int8 L_byte = 0;
#asm
//clrf H_byte
//clrf L_byte
movf mulcnd,W // move the multiplicand to W reg.
bcf STATUS,0 // Clear the carry bit in the status Reg.
btfsc mulplr,0
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc mulplr,1
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc mulplr,2
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc mulplr,3
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc mulplr,4
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc mulplr,5
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc mulplr,6
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc mulplr,7
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
#endasm
return H_byte*256+L_byte;
} |
Last edited by anestho on Sun Sep 09, 2007 7:08 pm; edited 1 time in total |
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anestho
Joined: 27 Dec 2006
Posts: 28
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| Posted: Sun Sep 09, 2007 7:07 pm |
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I eliminated the clrf for the H_byte and L_byte since I clear them above, and its duplicate code in .lst
I also reduced the instruction count by using W for one of the arguments.
CAN someone with a UART connection please test this routine for me? I only have 12F683 and it doesn't have a built in UART.
| Code: | / Begin Multiplier Routine
int16 mult8x8(int8 mulcnd, int8 W) // 35 instruction cycles = 0.5 x 35 = 17us
{
int8 H_byte = 0;
int8 L_byte = 0;
#asm
//clrf H_byte // removed, not needed
//cllrf L_byte // removed, not needed
//movf mulcnd,W // removed, not needed
bcf STATUS,0 // Clear carry bit
btfsc W,0 // compare W 0 bit with 0, if 1 goes to next line
addwf H_byte,1 // add 1 to MSB
rrf H_byte,1
rrf L_byte,1
btfsc W,1 // compare W 1 bit with 0, if 1 goes to next line
addwf H_byte,1 // add 1 to MSB
rrf H_byte,1
rrf L_byte,1
btfsc W,2
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc W,3
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc W,4
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc W,5
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc W,6
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
btfsc W,7
addwf H_byte,1
rrf H_byte,1
rrf L_byte,1
#endasm
return H_byte*256+L_byte;
}
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Also, how do I handle an interrupt while in this function? My interrupt uses the W register. Should I turn off the interrupt prior to entering this routine? |
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ckielstra
Joined: 18 Mar 2004
Posts: 3680
Location: The Netherlands
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| Posted: Mon Sep 10, 2007 1:18 am |
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Your code doesn't work because the function parameter with the name W is not the same as register W.
| Quote: | | CAN someone with a UART connection please test this routine for me? I only have 12F683 and it doesn't have a built in UART. |
You can test it yourself using the simulator in MPLAB.
| Quote: | | Also, how do I handle an interrupt while in this function? My interrupt uses the W register. Should I turn off the interrupt prior to entering this routine? |
It is the interrupt routine's responsibility for not changing any registers. You don't have to do anything. |
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Guest
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| Posted: Wed Oct 10, 2007 10:39 am |
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| I am not getting any answer. All variables are assigned zero when the fucntion is accessed. So how do I fix this? use movwf instead of movf? |
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Guest
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| Posted: Wed Oct 10, 2007 11:03 am |
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| Is there a way to do this in CCS C instead of asm? |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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| Posted: Wed Oct 10, 2007 11:53 am |
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Yes, there is. Use the _mul() function.
Here is the description of the _mul() function from the CCS manual:
| Quote: |
Performs an optimized multiplication. By accepting a different type than it
returns, this function avoids the overhead of converting the parameters
to a larger type. |
| Quote: | #include <16F877.H>
#fuses HS, NOWDT, NOPROTECT, BROWNOUT, PUT, NOLVP
#use delay(clock = 20000000)
#use rs232(baud=9600, xmit=PIN_C6, rcv=PIN_C7, ERRORS)
//======================================
void main()
{
int16 result;
int8 a, b;
a = 100;
b = 234;
result = _mul(a, b);
printf("result = %lu \n\r", result);
while(1);
} |
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