More likely a bug in mingw-w64:
#include <stdio.h>
#include <math.h>
volatile double x = 10.001000, y = -1.299000;
int main(){
int quo;
double rem = remquo(x, y, &quo);
printf("rem = %f, quo = %d\n", rem, quo);
}
With mingw-w64 this program gives the following output:
E:\Desktop>gcc test.c
E:\Desktop>a
rem = -0.391000, quo = 0
However, according to ISO C11 draft:
> 7.12.10.3 The remquo functions
> 2 The remquo functions compute the same remainder as the remainder functions.
> In
> the object pointed to by quo they store a value whose sign is the sign of x/y
> and whose
> magnitude is congruent modulo 2n to the magnitude of the integral quotient of
> x/y, where
> n is an implementation-defined integer greater than or equal to 3.
the value stored in `quo` must have the same sign with `x/y`.
In the above example, since `x/y`, which is about -7.699, is negative,
returning a non-negative value (zero in the above example) should be a bug.
------------------
Best regards,
lh_mouse
2016-09-07
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