Question

Is it possible to write a template that changes behavior depending on if a certain member function is defined on a class?

Here’s a simple example of what I would want to write:

template<class T> std::string optionalToString(T* obj) {     if (FUNCTION_EXISTS(T->toString))         return obj->toString();     else         return 'toString not defined'; } 

So, if class T has toString() defined, then it uses it; otherwise, it doesn’t. The magical part that I don’t know how to do is the ‘FUNCTION_EXISTS’ part.

Answer 1

Yes, with SFINAE you can check if a given class does provide a certain method. Here’s the working code:

#include <iostream>  struct Hello {     int helloworld() { return 0; } };  struct Generic {};      // SFINAE test template <typename T> class has_helloworld {     typedef char one;     struct two { char x[2]; };      template <typename C> static one test( decltype(&C::helloworld) ) ;     template <typename C> static two test(...);      public:     enum { value = sizeof(test<T>(0)) == sizeof(char) }; };      int main(int argc, char *argv[]) {     std::cout << has_helloworld<Hello>::value << std::endl;     std::cout << has_helloworld<Generic>::value << std::endl;     return 0; } 

I’ve just tested it with Linux and gcc 4.1/4.3. I don’t know if it’s portable to other platforms running different compilers.