Why does std :: pow float and int cause this overload?

Question

Following:

#include <cmath>

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

Raises warning a -Wconversionif it is enabled. Wandbox

It seems that doubleoverload is being called std::powwhere I expect overload to be selected float(with an exponent intother than float). Can someone who knows its overload explain why?

Answer 1

++ 11, pow - , double. double, , long double, long double.

[c.math]

++ float long double overloaded .

, , :

• , , long double, , , long double.

• , , , double integer, , , double.

• , , float.

, :

• long double → long double
• float → float
• → double

Answer 2

cpp reference

7) , 1-3). - , . , Promoted double, .

" - , " , double float.

float       pow( float base, float exp );