在c ++模板元编程中检测void方法

Question

我正在尝试编写模板元函数来检测类型是否具有void类型的成员函数。

目前我能够使用SFINAE来检测成员函数是否具有明确的类型，例如double，int等

 template<typename C> static auto Test(void*) -> decltype(int{std::declval<C>().foo()}, Yes{});

当然我可以反转它（如附带的代码片段所示）来测试它不是int，但我无法弄清楚如何测试它是无效的。

下面的代码片段目前输出

A  does not have void foo
B  has void foo
C  has void foo

但是，C的foo（）方法具有double类型，因此这是不正确的输出。 如何调整它以正确检查void foo() ？

#include <iostream>
#include <memory>

class A {
public:
    int foo() {
        return 0;
    }
};

class B {
public:
    void foo() {
    }
};

class C {
public:
    double foo() {
        return 0;
    }
};

template <typename T>
class has_void_foo {
private:
  typedef char Yes;
  typedef Yes No[2];

  template<typename C> static auto Test(void*) -> decltype(int{std::declval<C>().foo()}, Yes{});
  template<typename> static No& Test(...);

public:
    static bool const value = sizeof(Test<T>(0)) != sizeof(Yes);
};

int main(void) {
    std::cout << "A ";
    if (has_void_foo<A>::value) {
        std::cout << " has void foo";
    } else {
        std::cout << " does not have void foo";
    }
    std::cout << std::endl << "B ";
    if (has_void_foo<B>::value) {
        std::cout << " has void foo";
    } else {
        std::cout << " does not have void foo";
    }
    std::cout << std::endl << "C ";
    if (has_void_foo<C>::value) {
        std::cout << " has void foo";
    } else {
        std::cout << " does not have void foo";
    }
    std::cout << std::endl;

    return 0;
}

Answer 1

它遵循基于constexpr函数的可能解决方案：

#include <type_traits>

struct A {
    int foo() {
        return 0;
    }
};

struct B {
    void foo() {
    }
};

struct C {
    double foo() {
        return 0;
    }
};

template<typename T, typename R, typename... Args>
constexpr bool has_void_foo(R(T::*)(Args...)) { return std::is_void<R>::value; }

int main() {
    static_assert(not has_void_foo(&A::foo), "!");
    static_assert(has_void_foo(&B::foo), "!");
    static_assert(not has_void_foo(&C::foo), "!");
}