struct Thing {
    int foo(double,bool) {return 0;}
};

在编译期间有int foo(double,bool)成员函数.有很多方法可以做到这一点,大多数只是其他方式的变体.有人会想到一种与我在这里提到的5种方式截然不同(或至少相当有创意)的方式吗？我只是想学习一些模板和SFINAE的新技术.

#include <iostream>
#include <type_traits>

// Using void_t (this includes using std::is_detected).
template <typename T>
using void_t = void;

template <typename T,typename = void>
struct has_foo : std::false_type {};

template <typename T>
struct has_foo<T,void_t<decltype(static_cast<int>(std::declval<T>().foo(double{},bool{})))>
    > : std::true_type {};

// Using the ... default argument.
template <typename T>
struct hasfoo {
    template <typename U>
    static std::true_type test (decltype(static_cast<int(T::*)(double,bool)>(&T::foo))*);  // or 'decltype(static_cast<int>(std::declval<U>().foo(double{},bool{})))*' works fine too.

    template <typename>
    static std::false_type test (...);

    static constexpr bool value = decltype(test<T>(nullptr))::value;
};

// Overloads and trailing return types.
template <typename>
struct Helper : std::true_type {};

template <typename T>
auto helper(int) -> Helper<decltype(static_cast<int>(std::declval<T>().foo(double{},bool{})))>;

template <typename>
std::false_type helper(long);

template <typename T>
constexpr bool hasFoo() {return decltype(helper<T>(0))::value;}

// Comma operator (basically the same as the above).
template <typename T>
auto check(int) -> decltype(static_cast<int>(std::declval<T>().foo(double{},bool{})),std::true_type{});

template <typename T>
std::false_type check(...);

template <typename T>
using HasFoo = decltype(check<T>(0));

// Member function pointer template parameter.
template <typename T>
struct Hasfoo {
    template <typename U,int(U::*)(double,bool)>
    struct Tag;

    template <typename U>
    static constexpr bool test (Tag<U,&U::foo>*) {return true;}

    template <typename>
    static constexpr bool test (...) {return false;}

    static constexpr bool value = test<T>(nullptr);
};

// Tests
struct Thing {
    int foo(double,bool) {return 0;}
};

int main() {
    static_assert (has_foo<Thing>::value,"");
    static_assert (hasfoo<Thing>::value,"");
    static_assert (hasFoo<Thing>(),"");
    static_assert (HasFoo<Thing>::value,"");
}

编辑：我刚刚想起了一个优雅而更通用的解决方案,Yakk很久以前就提出了一个不同的问题(这是他的实际打字,只是为了匹配foo函数而修改)：

namespace Meta {
  namespace details {
    template<template<class...>class Z,class=void,class...Ts>
    struct can_apply : std::false_type {};
    template<template<class...>class Z,class...Ts>
    struct can_apply<Z,decltype((void)(std::declval<Z<Ts...>>())),Ts...>:
      std::true_type
    {};
  }
  template<template<class...>class Z,class...Ts>
  using can_apply = details::can_apply<Z,void,Ts...>;
}

template<class T>
using member_foo = decltype(static_cast<int(T::*)(double,bool)>(&T::foo));

template<class T>
using has_member_foo = Meta::can_apply<member_foo,T>;