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multiply problem

 
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pfournier



Joined: 30 Sep 2003
Posts: 89

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multiply problem
PostPosted: Mon Mar 06, 2006 11:06 am     Reply with quote

I am using PCWH 3.245
I tried using a simple multiply and found the answer was wrong.

1b9*4e2=6a52, should be 0x86952

I then discovered _mul(a,b). That worked fine.

Here is my demo code.
Why doesn't the simple multiply work right?
Did I do something wrong?

I tried searching the group first but I didn't see this.

Thanks, Pete



Code:

#define A 0x1b9
#define B 0x4e2

unsigned int16 mun16_A, mun16_B;
unsigned int32 mun32_A;

mun16_A=A;

printf("A=0x%lx B=0x%lx C=0x%lx \n\r",A,B,C);
printf("\n\r");
printf("mun16_A=0x%lx mun16_B=0x%lx mun16_C=0x%lx \n\r",mun16_A,mun16_B,mun16_C);
printf("mun32_A=0x%lx mun32_B=0x%lx mun32_C=0x%lx \n\r",mun32_A,mun32_B,mun32_C);
printf("\n\r");
printf("\n\r");
mun32_A=(int32_t)(mun16_A*B);
printf("should be 0x00086952, mun16_A*B=0x%lx\n\r",mun32_A);
mun16_B=_mul(mun16_A, B);
printf("should be 0x00086952, _mul(mun16_A,B)=0x%lx\n\r",mun16_B);

_________________
-Pete
pfournier



Joined: 30 Sep 2003
Posts: 89

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PostPosted: Mon Mar 06, 2006 11:20 am     Reply with quote

-oops
I took out some extra code an put in a mistake, here is the right stuff

Code:

#define A 0x1b9
#define B 1250
#define C 0x3ff
#define D 0x400

unsigned int16 mun16_A, mun16_B;
unsigned int32 mun32_A;

mun16_A=A;

printf("A=0x%lx B=0x%lx C=0x%lx \n\r",A,B,C);
printf("\n\r");
printf("mun16_A=0x%lx mun16_B=0x%lx\n\r",mun16_A,mun16_B);
printf("mun32_A=0x%lx\n\r",mun32_A);
printf("\n\r");
printf("\n\r");
mun32_A=(int32_t)(mun16_A*B);
printf("should be 0x00086952, mun16_A*B=0x%lx\n\r",mun32_A);
mun32_A=_mul(mun16_A, B);
printf("should be 0x00086952, _mul(mun16_A,B)=0x%lx\n\r",mun32_A);

_________________
-Pete
PCM programmer



Joined: 06 Sep 2003
Posts: 20404

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PostPosted: Mon Mar 06, 2006 12:57 pm     Reply with quote

Quote:
mun32_A=(int32_t)(mun16_A*B);

Before I look at your program, where's the typedef statement
for that special data type ?
pfournier



Joined: 30 Sep 2003
Posts: 89

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PostPosted: Mon Mar 06, 2006 1:23 pm     Reply with quote

Sorry, just change that to int32.

I have a stdint.h file that I use to make int8_t, int16_t and int32_t available cross platform. At one time I wrote c code that ran on a Rabbit processor and a PIC. It avoid the issue of an int being 8 bits on a PIC and 16 bits on a rabbit.
_________________
-Pete
PCM programmer



Joined: 06 Sep 2003
Posts: 20404

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PostPosted: Mon Mar 06, 2006 1:48 pm     Reply with quote

Fix it by casting one of the operands to 32-bits. See the changes
shown in bold below.
Quote:
mun32_A = (int32)mun16_A * B;
printf("should be 0x00086952, mun16_A * B = %lx\n\r", mun32_A);

mun32_A = _mul((int32)mun16_A, B);
printf("should be 0x00086952, _mul(mun16_A, B) = %lx\n\r", mun32_A);
pfournier



Joined: 30 Sep 2003
Posts: 89

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PostPosted: Mon Mar 06, 2006 4:21 pm     Reply with quote

Yup, that did it. Didn't need to cast in the _mul() though.
I thought just casting the whole operation would do it.
One of those things I don't do very often, so I never learned that lesson till now.

Thank you.
_________________
-Pete
Ttelmah
Guest







PostPosted: Mon Mar 06, 2006 4:28 pm     Reply with quote

This is the whole 'point' of using _mul. It takes two int_16 values, and multiplies them to produce an int32 result. Quicker and smaller than casting. When you casted the whole command in brackets, the effect was to multiply two int16 values, which therefore used int16 arithmetic, and then convert the result to int32 (which would be the default behaviour since the result was being transferred to an int32).

Best Wishes
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