Why zero times infinity = negative nan?

Asked May 25 '23 at 11:01

Active May 25 '23 at 11:10

Viewed 97 times

I have this code:

#include <iostream>

int main()
{
    double A = 1e3000000;
    double B = 0;
    double C = A * B;

    std::cout << "A = " << A << "\n";
    std::cout << "B = " << B << "\n";
    std::cout << "C = " << C << "\n";
}

Output:

A = inf
B = 0
C = -nan

Why is C negative? I tested this with clang 17 and also with online C++ compiler.

asked May 25 '23 at 11:01

green lantern

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2

https://stackoverflow.com/questions/21349847/positive-vs-negative-nans – interjay May 25 '23 at 11:08
5

fun fact: compiler dependant: https://godbolt.org/z/h8sjffh9K – ChrisB May 25 '23 at 11:13
It's weird and (I'd agree) somewhat surprising, but it's not wrong, because the sign bit has no meaning for NaNs. – Steve Summit May 25 '23 at 11:39
`A * B` can be done in higher precision in an optimizing compiler, resulting in different results – phuclv May 25 '23 at 14:42

Why zero times infinity = negative nan?

0 Answers0