C++ SFINAE enable_if_t in member function, how to disambiguate?

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Suppose we have some SFINAE member function:

class foo{     template <class S, class = std::enable_if_t<std::is_integral<S>::value, S>     void bar(S&& s);     template <class S, class = std::enable_if_t<!std::is_integral<S>::value, S>     void bar(S&& s); } 

If we declared it as above, then how can we define them? Both of their function signatures would look like:

template <class S, class> inline void foo::bar(S&& s){ ... do something ... } 

I have seen examples where one returns an std::enable_if_t<...> like:

template <class S, class> auto bar(S&& s) -> std::enable_if_t<!std::is_integral<S>::value, S>(...){     ... do something ... } 

To disambiguate based off of the return type. But I don't want to return anything.


since default arguments are not part of a function signature, make them not default

class foo{     template <class S, typename std::enable_if<std::is_integral<S>::value, int>::type = 0>     void bar(S&& s);     template <class S, typename std::enable_if<!std::is_integral<S>::value, int>::type = 0>     void bar(S&& s); }; 

Live Demo

EDIT: by popular demand, Here's the same code in C++17:

class foo{ public:     template <class S>     void bar(S&& s)     {         if constexpr(std::is_integral_v<S>)             std::cout << "is integral/n";         else             std::cout << "NOT integral/n";     } }; 

constexpr if statements are special to the compiler because the branch is chosen at compile time, and the non-taken branch isn't even instantiated

C++17 Demo


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