Is there provision specifying the restriction on "return type" in a function template expl-CodePudding

template<class T>
void fun(T){}

template<>
int fun(int){return 0;}

Consider this example, it is rejected by all implementations. However, I haven't found any persuasive provision in the current standard that specifies this explicit specialization declaration is ill-fomred. If exists, what is the rule?

In addition, the potential relevant rule may be that [temp.deduct.decl#2]

If, for the set of function templates so considered, there is either no match or more than one match after partial ordering has been considered ([temp.func.order]), deduction fails and, in the declaration cases, the program is ill-formed.

I think the meaning of "match" is not sufficiently clear here since "match" didn't clearly define anything.

CodePudding user response：

Your template definition did not match, as void fun(T) is not T fun(T) in your specialization.

You simply have to change to:

template<class T>
T fun(T){}

template<>
int fun(int){}

BTW: All this results in a lot of warning because you did not return anything :-)

CodePudding user response：

You were not returning anything in your int specialization. And that is illformed code.

The correct syntax would be this :

#include <iostream>

template<class T>
auto fun(T value) 
{
    std::cout << value << "\n";
}

template<>
auto fun<int>(int value) 
{
    auto retval = value   2;
    std::cout << retval << "\n";
    return retval;
}

int main()
{
    fun("hello");
    int value = fun(40);

    return value;
}