🚀 go-pugleaf

In article <1ghtanINNcf9@cs.tut.fi> pjh@cs.tut.fi (Haavisto Petri) writes:

>I don't have the standard, but it seems clear to me that given an
>expression of type double with a positive value, the following
>should always be true:
>
>		(long) (expression) == (long) floor(expression)

I also don't have the standard, but I have observed that the above will
fail for cases where the expression is negative.  Both (long) and (int) 
round toward zero, not negative infinity.  I wrote the following test
program and tried it on both a SUN SparkStation and an IBM RS/6000 and 
got the same results:

#include <math.h>
#include <stdio.h>

main(argc, argv)
int argc;
char **argv;
{
    double x;

    if (argc > 1) {
        x = atof(argv[1]);

        printf ("the number is %lf, (long) of it is %d, floor of it is %d\n", x, (long)x, (long) floor(x));
        if ((long) x == (long) floor (x))
            printf ("they are equal\n");
        else
            printf ("they are not equal\n");
    }
    else {
        printf ("usage:  %s n\nwhere n is a (double) number\n", argv[0]);
    }

    exit (0);
}

Gordon