template class restriction

Question

I\'m wondering if there is any way to restrict generating code for a template using custom conditions in my case i want to function foo to be called only if template class T has

Answer 1

Yes, following technique can be used (for public inheritance). It will cause an overhead of just one pointer initialization.

Edit: Re-writing

template<typename Parent, typename Child>
struct IsParentChild
{
  static Parent* Check (Child *p) { return p; }
  Parent* (*t_)(Child*);
  IsParentChild() : t_(&Check) {} // function instantiation only
};

template<typename T>
void foo ()
{
  IsParentChild<Bar, T> check;
// ...
}

Answer 2

You can restrict T though using "Substitution Failure Is Not An Error" (SFINAE):

template <typename T>
typename std::enable_if<std::is_base_of<bar, T>::value>::type foo()
{

}

If T is not derived from bar, specialization of the function template will fail and it will not be considered during overload resolution. std::enable_if and std::is_base_of are new components of the C++ Standard Library added in the forthcoming revision, C++0x. If your compiler/Standard Library implementation don't yet support them, you can also find them in C++ TR1 or Boost.TypeTraits.