在模板专业化和重载C ++上没有得到预期的歧义

Question

Consider

Class Wow{
    public:
        //main metod
        template<typename T>
        void foo(T t){
            cout << t << endl;
        }

        template<>
        void foo<int>(int t){
            cout << "specialization" << endl;
        }

        void foo(int t){
            cout << "overloading" << endl;
        }
}

and the main is

Wow wow;
wow.foo(2.2);
wow.foo(1);

this outputs

2.2
overloading

My question is why does that even compile? practically, foo is defined twice as void foo(int) .

1) why does this pass?

2) why does the compiler choose the overloading one?

Thanks

Answer 1

1) Because there is no problem here. There is template function, function template specialization and overloading. You can call template specialization like this:

wow.foo<int>(3);

2) Overload has better match, than template specialization, if compiler can call this function with arg.

n4926 13.3.3/1.7

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 not a function template specialization and F2 is a function template specialization