Floating-Point Accuracy – Why Double-Precision Operations Fail

My other answer was dead wrong.

#include <stdio.h>

int
main(int argc, char *argv[])
{
    unsigned long long n = 1ULL << 53;
    unsigned long long a = 2*(n-1);
    unsigned long long b = 2*(n-2);
    printf("%llu\n%llu\n%d\n", a, b, (double)a == (double)b);
    return 0;
}

Compile and run to see:

18014398509481982
18014398509481980
0

a and b are just 2*(2⁵³-1) and 2*(2⁵³-2).

Those are 17-digit base-10 numbers. When rounded to 16 digits, they are the same. Yet a and b clearly only need 53 bits of precision to represent in base-2. So if you take a and b and cast them to double, you get your counter-example.

The specific reason in your case is that the real number 0.94 cannot be represented exactly in a double precision floating point. When you type 0.94, the actual number stored is 0.939999999999999946709294817992486059665679931640625.