Check if parameter pack contains a type

问题

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 };
};

回答1:

No, you have to use (partial) specialization with variadic templates to do compile-time computations like this:

#include <type_traits>

template < typename Tp, typename... List >
struct contains : std::true_type {};

template < typename Tp, typename Head, typename... Rest >
struct contains<Tp, Head, Rest...>
: std::conditional< std::is_same<Tp, Head>::value,
    std::true_type,
    contains<Tp, Rest...>
>::type {};

template < typename Tp >
struct contains<Tp> : std::false_type {};

There is only one other intrinsic operation for variadic templates and that is the special form of the sizeof operator which computes the length of the parameter list e.g.:

template < typename... Types >
struct typelist_len
{
   const static size_t value = sizeof...(Types);
};

Where are you getting "it has serious compilation-time overhead" with boost mpl from? I hope you are not just making assumptions here. Boost mpl uses techniques such as lazy template instantiation to try and reduce compile-times instead of exploding like naive template meta-programming does.

回答2:

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_prsent {
    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 even 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.

回答3:

If you want to avoid manual type recursion, std::common_type appears to me to be the only utility in the STL which is a variadic template, and hence the only one which could potentially encapsulate recursion.

Solution 1

std::common_type finds the least-derived type in a set of types. If we identify numbers with types, specifically high numbers with less-derived types, it finds the greatest number in a set. Then, we have to map equality to the key type onto a level of derivation.

using namespace std;

struct base_one { enum { value = 1 }; };
struct derived_zero : base_one { enum { value = 0 }; };

template< typename A, typename B >
struct type_equal {
 typedef derived_zero type;
};

template< typename A >
struct type_equal< A, A > {
 typedef base_one type;
};

template< typename Key, typename ... Types >
struct pack_any {
 enum { value =
     common_type< typename type_equal< Key, Types >::type ... >::type::value };
};

Solution 2

We can hack common_type a little more. The standard says

A program may specialize this trait if at least one template parameter in the specialization is a user-defined type.

and describes exactly what is inside it: a recursive partial specialization case, a case which applies a binary operator, and a terminal case. Essentially, it's a generic fold function, and you can add whatever binary operation you please. Here I used addition because it's more informative than OR. Note that is_same returns an integral_constant.

template< typename Addend >
struct type_sum { // need to define a dummy type to turn common_type into a sum
    typedef Addend type;
};

namespace std { // allowed to specialize this particular template
template< typename LHS, typename RHS >
struct common_type< type_sum< LHS >, type_sum< RHS > > {
    typedef type_sum< integral_constant< int,
     LHS::type::value + RHS::type::value > > type; // <= addition here
};
}

template< typename Key, typename ... Types >
struct pack_count : integral_constant< int,
 common_type< type_sum< is_same< Key, Types > > ... >::type::type::value > {};

来源：https://stackoverflow.com/questions/2118541/check-if-parameter-pack-contains-a-type