Templates: only execute a method if it has a class

Question

I want to write a function that executes a method of some template class, but should also compile fine if the class does not have this. In this case, it should just not call the function.

struct A { void func() {} }; struct B { }; template <typename T> void anotherFunc(T t) { //do t.func() here if T implements func, just do nothing if it doesn't. } 

Is it possible somehow?

Answer 1

 // type_sink takes a type, and discards it. type_sink_t is a C++1y style using alias for it template<typename T> struct type_sink { typedef void type; }; template<typename T> using type_sink_t = typename type_sink<T>::type; // has_func is a traits class that inherits from `true_type` iff the expression t.func() // is a valid one. `std::true_type` has `::value=true`, and is a good canonical way to // represent a compile-time `bool`ean value. template<typename T,typename=void> struct has_func : std::false_type {}; template<typename T> struct has_func< T, type_sink_t< decltype( std::declval<T&>().func() ) > > : std::true_type {}; // helpers for tag dispatching. namespace helper_ns { template<typename T> void anotherFunc( T&& t, std::false_type /* has_func */ ) {} template<typename T> void anotherFunc( T&& t, std::true_type /* has_func */ ) { std::forward<T>(t).func(); } } // take the type T, determine if it has a .func() method. Then tag dispatch // to the correct implementation: template<typename T> void anotherFunc(T t) { helper_ns::anotherFunc( std::forward<T>(t), has_func<T>() ); } 

is a C ++ 11 solution that sends a tag to a feature class that determines if t.func() valid expression.