Question

I was looking at the implementation of the is_class template in Boost, and ran into some syntax I can’t easily decipher.

    template <class U> static ::boost::type_traits::yes_type is_class_tester(void(U::*)(void));
    template <class U> static ::boost::type_traits::no_type is_class_tester(...);

How do I interpret void(U::*)(void) above? I’m familiar with C, so it appears somewhat analogous to void(*)(void), but I don’t understand how U:: modifies the pointer. Can anyone help?

Thanks

Answer 1

* indicates a pointer, because you can access its contents by writing *p. U::* indicates a pointer to a member of class U. You can access its contents by writing u.*p or pu->*p (where u is an instance of U).

So, in your example, void (U::*)(void) is a pointer to a member of U that is a function taking no arguments and returning no value.

Example:

class C { void foo() {} };

typedef void (C::*c_func_ptr)(void);

c_func_ptr myPointer = &C::foo;