c++ - Check if parameter pack contains a type

Question

I was wondering if C++0x provides any built-in capabilities to check if a parameter pack of a variadic template contains a specific type. Today, boost:::mpl::contains can be used to accomplish this if you are using boost::mpl::vector as a substitute for variadic templates proper. However, it has serious compilation-time overhead. I suppose, C++0x has compiler-level support for std::is_same. So I was thinking if a generalization like below is also supported in the compiler.

template <typename... Args, typename What>
struct is_present
{
  enum { value = (What in Args...)? 1 : 0 };
};

Answer 1

Fortunately, the C++ standard has evolved. With C++1z aka C++17, you can finally iterate easily over parameter packs. So the code for the answer is (almost) as simple, as suggested in the question:

template<typename What, typename ... Args>
struct is_present {
    static constexpr bool value {(std::is_same_v<What, Args> || ...)};
};

The weird-looking (std::is_same_v<What, Args> || ...) is expanded by the compiler internally to (std::is_same_v<What, Args[0]> || std::is_same_v<What, Args[1]> || ...), which is exactly, what you want. It even correctly yields false with an empty Args parameter pack.

It is even possible to do the whole check inline in a function or method - no helper structs are required anymore:

template<typename T, typename ... List>
void foo(T t, List ... lst)
{
    if constexpr((std::is_same_v<T, List> || ...)) {
        std::cout << "T is in List" << std::endl;
    } else {
        std::cout << "T is not in List" << std::endl;
    }
}

Note: This has been taken from another question, that was marked as a duplicate of this question. As this is the "canonical" question for this topic, I added that important information here.