函数重载和模板

Question

#include <iostream>
using namespace std;


template <typename ReturnType, typename ArgumentType>
ReturnType Foo(ArgumentType arg){}


template <typename ArgumentType>
string Foo(ArgumentType arg) { cout<<"inside return 1"<<endl; return "Return1"; }

int main(int argc, char *argv[])
{

    Foo<int>(2);        
    return 0;
}

以上代码抛出如下错误。

In function 'int main(int, char**)': 
34:18: error: call of overloaded 'Foo(int)' is ambiguous 
34:18: note: candidates are: 
7:12: note: ReturnType Foo(ArgumentType) [with ReturnType = int; ArgumentType = int] 
20:8: note: std::string Foo(ArgumentType) [with ArgumentType = int; std::string = std::basic_string<char>]

因为，函数重载只考虑函数名、参数类型列表和封闭的命名空间。为什么会抛出这个错误？

Answer 1

对Foo<int>(2) 的调用可以是：

ReturnType Foo(ArgumentType arg); //with ReturnType = int, ArgumentType deduced as int

或

string Foo(ArgumentType arg); //with ArgumentType = int

编译器无法判断你想要哪一个，因为它们都有相同的参数：

int Foo(int);
string Foo(int);