Question

How do I say:

template<typename T>
class X {
  // if T has method x(), define
  //   Y x() { return t.x() }

  T t;
};

Answer 1

Just define it.

If X::x isn’t called, then T::x doesn’t have to exist either. If X::x is called and T::x doesn’t exist, the error message will point to the use of X::x. Most compilers would use wording along the lines of: “Unknown identifier x while compiling Y X<Something>::x(void) within this context: whatever called X::x() for a Something that doesn’t support it“.

EDIT: Since you’re using C++0x, by all means use decltype:

template<typename T>
class Forwards {
     T t;
public:
    decltype(this->t.x()) x() { return this->t.x(); }
};

I’m not 100% sure about whether to use decltype(T::x()), decltype(t.x()), or decltype(this->t.x()), but I’m pretty sure this should work. If t doesn’t supply x, then the Forwards::x() function wouldn’t be able to be instantiated. This still isn’t perfect forwarding, since you need to know the argument list a-priori, but now you can deal with return type variation.