When does NaN not propagate in C++?

Question

NaN propagates through "most" operations as described in NaN - Wikipedia.

I would like to know the operations which NaN does NOT propagate. For example, I'm coding in C++ and found that following code prints 1, which is not NaN.

const double result = std::pow(1, std::numeric_limits<double>::quiet_NaN());
std::cout << result << std::endl;

For std::powfunction, this behavior is described in std::pow - cppreference.com.

Could you share any other examples?

Answer 1

Here's an example demonstrating functions of NaN which return non-NaN. The list is in IEEE 754-2008, 9.2.1 Special values (there are some others functions, but they don't seem to be implemented in C++):

#include <cmath>
#include <limits>
#include <iostream>

int main()
{
    const auto nan=std::numeric_limits<double>::quiet_NaN();
    const auto inf=std::numeric_limits<double>::infinity();
    std::cout << std::hypot(nan,inf) << '\n';
    std::cout << std::hypot(inf,nan) << '\n';
    std::cout << std::pow(nan, 0) << '\n';
    std::cout << std::pow(1,nan) << '\n';
}

The output is:

inf
inf
1
1