Compile time type checking C++

Question

I have created a type list. I then create a class using a template passing the type list. When I call the print function of the class with a some types not specified they ar

Answer 1

You would have to make your print function into a template and then check whether the types match:

template <typename U>
void print(const U & u)
{
  // use std::is_same<typename std::decay<T::Head>::type, typename std::decay<U>::type>::value
}

Here I'm stealing is_same and decay from <type_traits>, but if you don't have C++11, you can either take them from TR1 or from Boost, or just write them yourself, as they're very simple type modifier classes.

The conditional would best go into a static_assert, which is another C++11 feature, but there exist similar constructions for C++98/03 that produce a compile-time error under a certain condition.

Answer 2

Add a private function template

template<typename T> void print(T);

which doesn't need an implementation. This should catch all types for which no explicit print exists, and since it is private, it will give an error message.

Answer 3

You could take your arguments by non-const reference, forcing them to be the exact same type. However you can no longer use it with const variables or literals.