Re: Modulo operation in C for -ve values

[Ingo Krabbe <ingo dot krabbe at eoa dot de>]

Am Freitag, 16. November 2007 10:31:35 schrieb Andrew Haley:
> I've redirected this to gcc-help, since it's a question about using
> gcc, not about developing gcc.
>
> Deepak Gaur writes:
>  > Subject: Modulo operation in C for -ve values
>  >
>  > The Modulo operation as specified in
>  > http://xenia.media.mit.edu/~bdenckla/thesis/texts/htthe/node13.html says
>  > that for a fraction like n/k which can be expressed as n/k = i + j/k the
>  > C division and mod operation should yeild
>  > n div k = i (integer part)
>  > n mod k = j (remainder part)
>  > For n +ve above is true
>  > For n -ve
>  > -n/k = -i + j/k
>  > -n div k = -i
>  > -n mod k = j (+ve remainder)

actually in this problem you can see the sense behind your thesis

It is statet there that it seems that  -3 ./. 7 = 0 + (-3/7)
where 

i = -((-n) div k)   and   j = - ((-n) mod k)   (n<0) [WRONG!]

does not hold but

i = -((-n) div k)-1 and   j = - ((-n) mod k)+k  (n<0)  [works]

moving the sign carrier to the integer part.

This always works since  (0<=j<k) !

Now you have your complete algorithm.

Happy mathing

bye ingo


>  >
>  > But running a sample program on Redhat enterprise Linux EL4
>  > with gcc version 3.4.3 20041212 (Red Hat 3.4.3-9.EL4)
>  > on a Intel PIV Machine
>  >
>  > #include <stdio.h>
>  > #include <stdlib.h>
>  > #include <math.h>
>  > int main()
>  > {
>  > int n,k,j;
>  > n=-3;
>  > k=8; /* k is power of 2 */
>  > j=(n/k);
>  > printf("n n div k = %d", j);
>  > j=(n%k);
>  > printf("\n n mod k = %d", j);
>  > j=(n) & (k-1);
>  > printf("\n n & k-1 = %d", j);
>  > }
>  > gives following output for n = -3 k = 8
>  > n div k = 0
>  > n mod k = -3
>  > n & k-1 = 5
>  > though it should have been as per hypothesis proposed in
>  > http://xenia.media.mit.edu/~bdenckla/thesis/texts/htthe/node13.html
>  > n div k = -1
>  > n mod k = 5
>  > n & k-1 = 5
>  >
>  > Which is correct(0,-3,5) or (-1,5,5)?
>
> gcc is correct.  Your mistake is to assume that the C operator %
> represents the mathematical modulo.  It doesn't: % is the remainder
> after division, not modulo arithmetic.  C only supports modulo on
> unsigned quantities.
>
> Andrew.



-- 

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I n g o   K r a b b e
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