c++ - Checking whether a function (not a method) exists in c++11 via templates -

so sfinae , c++11, possible implement 2 different template functions based on whether 1 of template parameters can substituted.

for example

struct boo{    void saysomething(){ cout << "boo!" << endl; } };  template<class x> void makeitdosomething(decltype(&x::saysomething), x x){    x.saysomething(); }  template<class x> void makeitsaysomething(int whatever, x x){     cout << "it can't anything!" << endl; }   int main(){    makeitsaysomething(3);    makeitsaysomething(boo()); } 

or along line.

my question is.. how 1 same thing, non-member functions?

in particular i'm trying check if there's such thing an:

operator<<(std::ostream& os, x& whateverclass); 

that exists. possible test it?

edit: question different : is possible write template check function's existence? in i'm trying see whether function exists, not method

i find void_t trick preferable traditional sfinae method shown in other answer.

template<class...> using void_t = void; // in c++17 working paper! // gcc <= 4.9 workaround: // template<class...> struct voider { using type = void; }; // template<class... t> using void_t = typename voider<t...>::type;  template<class, class = void> struct is_ostreamable : std::false_type {};  template<class t> struct is_ostreamable<t, void_t<decltype(std::declval<std::ostream&>() <<                                          std::declval<t>())>> : std::true_type {}; 

the partial specialization selected if , if expression well-formed.

demo. note & in std::declval<std::ostream&>() important, because otherwise std::declval<std::ostream>() rvalue , you'll ostream's catchall rvalue stream insertion operator, , report streamable.

the above code checks operator<< can accept t rvalue . if want check 1 accepts lvalue t, use std::declval<t&>().