为什么 float 和 int 的 std::pow 会调用此重载？

Question

以下：

#include <cmath>

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

在打开时触发 -Wconversion 警告。 Wandbox

似乎调用了 std::pow 的 double 重载，我希望选择 float 重载（将 int 指数转换为float)。知道他超载的人能解释一下为什么吗？

Answer 1

来自 cpp reference

7) A set of overloads or a function template for all combinations of arguments of arithmetic type not covered by 1-3). If any argument has integral type, it is cast to double. If any argument is long double, then the return type Promoted is also long double, otherwise the return type is always double.

"If any argument has integral type, it is cast to double" 然后它会调用，将你的双精度转换为浮点数。

float       pow( float base, float exp );

Answer 2

自 C++11 起，混合参数 pow 将任何整型参数转换为双精度。混合参数函数的 return 类型总是 double 除非一个参数是 long double 那么结果是 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.

所以总结一下：

一个参数 long double >> long double
两个参数 float >> float
否则>> double

为什么 float 和 int 的 std::pow 会调用此重载？

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

c++

overload-resolution

c++14

[c.math]