enable_if: minimal example for void member function with no arguments

问题

I am trying to get a better understanding of std::enable_if in C++11 and have been trying to write a minimal example: a class A with a member function void foo() that has different implementations based on the type T from the class template.
The below code gives the desired result, but I am not understanding it fully yet. Why does version V2 work, but not V1? Why is the "redundant" type U required?

#include <iostream>
#include <type_traits>

template <typename T>
class A {

    public:

        A(T x) : a_(x) {}

        // Enable this function if T == int
        /* V1 */ // template <           typename std::enable_if<std::is_same<T,int>::value,int>::type = 0>
        /* V2 */ template <typename U=T, typename std::enable_if<std::is_same<U,int>::value,int>::type = 0>
        void foo() { std::cout << "\nINT: " << a_ << "\n"; }

        // Enable this function if T == double
        template <typename U=T, typename std::enable_if<std::is_same<U,double>::value,int>::type = 0>
        void foo() { std::cout << "\nDOUBLE: " << a_ << "\n"; }

    private:

        T a_;

};

int main() {
    A<int> aInt(1); aInt.foo();
    A<double> aDouble(3.14); aDouble.foo();
    return 0;
}

Is there a better way to achieve the desired result, i.e. for having different implementations of a void foo() function based on a class template parameter?

回答1:

I know this wont fully answer your question, but it might give you some more ideas and understanding of how you can use std::enable_if.

You could replace your foo member functions with the following and have identical functionality:

template<typename U=T> typename std::enable_if<std::is_same<U,int>::value>::type
foo(){ /* enabled when T is type int */ }

template<typename U=T> typename std::enable_if<std::is_same<U,double>::value>::type
foo(){ /* enabled when T is type double */ }

A while back I gained a pretty good understanding of how enable_if works, but sadly I have forgotten most of its intricacies and just remember the more practical ways to use it.

回答2:

As for the first question: why V1 doesn't work? SFINAE applies only in overload resolution - V1 however causes error at the point where type A is instantiated, well before foo() overload resolution.

I suppose there are lot's of possible implementations - which is the most appropriate depends on an actual case in question. A common approach would be to defer the part of A that's different for different template types to a helper class.

template <typename T>
class A_Helper;

template <>
class A_Helper<int> {
public:
    static void foo( int value ){
        std::cout << "INT: " << value << std::endl;
    }
};

template <>
class A_Helper<double> {
public:
    static void foo( double value ){
        std::cout << "DOUBLE: " << value << std::endl;
    }
};

template <typename T>
class A {
public:

    A( T a ) : a_(a) 
    {}

    void foo(){
        A_Helper<T>::foo(a_);
    }

private:
    T a_;
};

The rest of A can be declared only once in a generic way - only the parts that differ are deferred to a helper. There is a lot of possible variations on that - depending on your requirements...

来源：https://stackoverflow.com/questions/41667905/enable-if-minimal-example-for-void-member-function-with-no-arguments

标签

c++

c++11

void

sfinae

enable-if