nils pipenbrinck
asked on
muldiv64 on small cpu's
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)
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)
>> x = (a/c)*b;
Which will give you an incorrect result due to premature rounding
Which will give you an incorrect result due to premature rounding
ASKER
sganter, as alexo already said.. this will give incorrect results.
nils
nils
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
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
(a/c)*(b/c)*c+a/c*(b%c)+b/
ASKER
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.)
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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x = (a/c)*b; /* Instead of x = (a*b)/c; */