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BadBrainBoy
Joined: 28 Apr 2005 Posts: 4

8x8 Multiply with 16bit Result 
Posted: Fri May 13, 2005 4:41 pm 


Hi All,
The default when doing int8 * int8 with a hardware multiplier PIC is to keep only the LSB of the result (PRODL). This is C standard, I suppose, but not very useful.
If you want the full 16bit result from int8 * int8, you can use this (written for the 18F2220):
=========================================
#byte PRODH = 0xff4 // HW multiply result MSB
#byte PRODL = 0xff3 // HW multiply result LSB
// MULT macro
// Performs 8x8 hardware multiply of x and y and stores the 16bit
// result in xy
#define MULT(x,y,xy) \
#asm \
movf x,w \
mulwf y \
#endasm \
xy = MAKE16(PRODH,PRODL);
=========================================
Now just write MULT(arg1,arg2,result) when you want to multiply two int8 variables named arg1 and arg2 and store the product in int16 result.
This is very efficient, it compiles to only 4 machine instructions with 6 instruction cycles. It is set up for straight unsigned multiplication; you may fiddle with it for signed.
Have fun! 


sd211
Joined: 15 Nov 2005 Posts: 4 Location: NC, USA

Re: 8x8 Multiply with 16bit Result 
Posted: Sun Jan 01, 2006 7:46 pm 


BadBrainBoy wrote:  Hi All,
The default when doing int8 * int8 with a hardware multiplier PIC is to keep only the LSB of the result (PRODL). This is C standard, I suppose, but not very useful.
If you want the full 16bit result from int8 * int8, you can use this (written for the 18F2220):
=========================================
#byte PRODH = 0xff4 // HW multiply result MSB
#byte PRODL = 0xff3 // HW multiply result LSB
// MULT macro
// Performs 8x8 hardware multiply of x and y and stores the 16bit
// result in xy
#define MULT(x,y,xy) \
#asm \
movf x,w \
mulwf y \
#endasm \
xy = MAKE16(PRODH,PRODL);
=========================================
Now just write MULT(arg1,arg2,result) when you want to multiply two int8 variables named arg1 and arg2 and store the product in int16 result.
This is very efficient, it compiles to only 4 machine instructions with 6 instruction cycles. It is set up for straight unsigned multiplication; you may fiddle with it for signed.
Have fun! 
There is function _mul(x,y), which will give you very similar code, maybe faster by 1 cycle.
Sergey 


Jimmy
Joined: 21 Aug 2006 Posts: 1


Posted: Mon Aug 21, 2006 3:03 am 


I tried c = _mul(arg1,arg2);. It's not working properly. I use PIC16F917 and simulate with MPLAB. The main function is like this.
void main()
{
int8 varA=50,varB=100;
long int c;
c = _mul(varA,varB);
while(TRUE);
}
in the watch window show c = 34835 insteal to show c = 5000.
I try many value, but always wrong result.
Is there anyone can help on this? thanks. 


Mircea
Joined: 12 Jan 2007 Posts: 7 Location: Europe


Posted: Tue May 06, 2008 3:10 pm 


Code:  Not all processors have mul statement.
Here is an efficent algorithm to multiply two
integers.
Consider two numbers, example a=122 and b=57.
We want to get the result of their multiplication.
The expected result is 122*57 = 6954.
We are using the "generic" formula:
2*a b
a * b =  * 
1 2
Multiplying a by 2 and dividing b by 2 (integer divison),
we get this table:

1 2 3

122 57 ()
244 28
488 14
976 7 ()
1952 3 ()
3904 1 () stop algorithm

The marker () in the third column means that the corresponding
value in column 2 is NOT divisible by 2.
Now we add the values in column 1 on marked rows:
122+
976
1952
3904

6954 (which is the good result).
One can observe:

1. The operations involved:
 multiplication by 2
 division by 2
 test bit 0 to see if it is odd
are easy to perform by controllers.
2. It is not necessary to store the partial
values in columns 1 and 2, everything cand be done
in one single loop.
3. The algorithm is very efficient.
Some peolpe said that the above algorithm is
used to create mul or imul elementary statements
inside the controllers or processors that have a such statement.






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