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muldiv64 on small cpu's

Posted on 1998-06-24
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Last Modified: 2008-02-01
Hi.

I have a small problem.

I need to calculate  x =  (a*b)/c.

Problem is, that a,b and c are 32 bit long, and the machine doesn't have a 64/32 bit division. The code the compiler generates in this case is ugly and slow (yep, it's time-critical). Does anyone has an idea how to do a fast version with 2 or maybe 4 ordinary 32/32 divisions?

Thank you in advance,

 Nils Pipenbrinck

(P.S. Please don't quote ugly c-code from the gnu lib)
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Question by:nils pipenbrinck
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by:sganta
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Hello !

You use
x = (a/c)*b;    /* Instead of x = (a*b)/c; */

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by:alexo
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>> x = (a/c)*b;
Which will give you an incorrect result due to premature rounding
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by:nils pipenbrinck
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sganter, as alexo already said.. this will give incorrect results.

nils
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by:sganta
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Hello !
x = (a/c)*b;    /* Instead of x = (a*b)/c; */
is correct because
First it will get the result for a/c since it is in the paranthesis.
i.e., 32bit/32 bit division, result will be 32 bit only

Then it will take the result and multiply with b.

The equivalent statement in the assembly language is like this
MOVE value_of_a to AX
DIV AX,value_of_c /* AX contains the RESULT of A/C */
MUL AX,value_of_b /*AX  contains the FINAL RESULT  */

You test this program. Here it works. If it does'nt work, I don't mind.
You can reject my answer
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by:ozo
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(a/c)*(b/c)*c+a/c*(b%c)+b/c*(a%c)+(a%c)*(b%c)/c
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by:nils pipenbrinck
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x = (a/c)*b;    /* Instead of x = (a*b)/c; */

 a = 1
 b = 1000
 c = 2

 x = (a/c)*b would give:

  (1/2) * 1000
  = 0 * 1000
  = 0

while my formula gives:

 x = (a*b)/c

 (1*1000)/2 = 500

your substitution won't give the correct results in this case.
anyways  we  found a fast way to solve this problem.

Thank you.

(just answer again, I reopend the question for 50 points. then 'll give you the 50 points for your work.)
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sganta earned 100 total points
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I am sorry, I will apreciate it.
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