IEEE 754: can casting before division cause loss of precision?

Question

Do there exist two integers i and j that both fit into an IEEE 754 double (are smaller than DBL_MAX), but such that to_double(i)/to_double(j) is not equal to to_double(i/j), where this i/j is performed with unlimited precision?

(We can assume to_double is round-to-even if that matters).

My question is similar to Invertability of IEEE 754 floating-point division, but I don't think it is equivalent, or at least I don't see how to use it to get a counter-example to my question.

Eric Postpischil · Accepted Answer · 2019-06-19T10:36:17.243

Yes. In a C implementation where double is IEEE-754 basic 64-bit binary floating-point (with 53-bit significands) and long double has 64-bit significands, the output of:

#include <stdio.h>

int main(void)
{
    long double x = 0x1p154L - 0x1p101L + 0x1p100L;
    long double y = 0x1p153L + 0x1p101L - 0x1p100L;
    long double z = x / y;
    double X = x;
    double Y = y;
    double Z = X/Y;
    printf("x = %La.\n", x);
    printf("y = %La.\n", y);
    printf("z = %La.\n", z);
    printf("X = %a.\n", X);
    printf("Y = %a.\n", Y);
    printf("Z = %a.\n", Z);
    printf("(double) z = %a.\n", (double) z);
}

is:

x = 0xf.ffffffffffffcp+150.
y = 0x8.0000000000004p+150.
z = 0xf.ffffffffffff4p-3.
X = 0x1p+154.
Y = 0x1p+153.
Z = 0x1p+1.
(double) z = 0x1.ffffffffffffep+0.

x / y is performed with long double precision, of course, rather than infinite precision, but it captures sufficient information to show the result with infinite precision would have the same end result—inserting #include <math.h> and z = nexttowardl(z, INFINITY); changes (double) z to be 0x1.fffffffffffffp+0, but this is still not equal to Z.

IEEE 754: can casting before division cause loss of precision?

1 Answers1