c++ - 当重载方法将模板类作为参数时会发生什么

Question

C++ 允许我们使用模板编写通用函数。并且它还具有函数重载的特性。

我编写了如下程序:

#include <iostream>

using namespace std;
template <typename T>
void test(T a)
{
    cout<<"using template";    
}
void test(int a)
{
    cout<<"using int";
}
int main()
{

   test(10);
   return 0;
}

结果是:

using int

我想知道选择特定方法的依据是什么？

Answer 1

非模板函数比函数模板更匹配。

引用(C++ 草案标准 N3337):

13.3.3 Best viable function

...

Given these definitions, a viable function F1 is defined to be a better function than another viable function F2 if for all arguments i, ICSi(F1) is not a worse conversion sequence than ICSi(F2), and then

...

— F1 is a non-template function and F2 is a function template specialization, or, if not that,

— F1 and F2 are function template specializations, and the function template for F1 is more specializedthan the template for F2 according to the partial ordering rules described in 14.5.6.2.