Why does std::pow of a float and int invoke this overload?

Question

The following:

#include <cmath>

int main()
{
    float base = 2.0f;
    float result = std::pow(base, 2);
    return 0;
}

Triggers a -Wconversion warning in case it is turned on. Wandbox

It seems that the double overload of std::pow is invoked, where I would expect the float overload to be chosen (with the int exponent being cast to float). Can someone who knows his overloads please explain why?

Answer 1

Since C++11, mixed argument pow has any integral argument cast to double. The return type of the mixed argument functions is always double except when one argument is long double then the result is long double.

[c.math]

In addition to the double versions of the math functions in , C++ adds float and long double overloaded versions of these functions, with the same semantics.

Moreover, there shall be additional overloads sufficient to ensure:

• If any argument corresponding to a double parameter has type long double, then all arguments corresponding to double parameters are effectively cast to long double.

• Otherwise, if any argument corresponding to a double parameter has type double or an integer type, then all arguments corresponding to double parameters are effectively cast to double.

• Otherwise, all arguments corresponding to double parameters are effectively cast to float.

So to sum up:

• One argument long double >> long double
• Both arguments float >> float
• Otherwise >> double