检查 class 是否有可能重载的函数调用运算符

Question

我想知道是否可以在 C++20 中实现一个特征来检查类型 T 是否具有可能的 overloaded/possibly 模板函数调用运算符： operator().

// Declaration
template <class T>
struct has_function_call_operator;

// Definition
???  

// Variable template
template <class T>
inline constexpr bool has_function_call_operator_v 
= has_function_call_operator<T>::value;

以便像下面这样的代码会导致正确的结果：

#include <iostream>
#include <type_traits>

struct no_function_call_operator {
};

struct one_function_call_operator {
    constexpr void operator()(int) noexcept;
};

struct overloaded_function_call_operator {
    constexpr void operator()(int) noexcept;
    constexpr void operator()(double) noexcept;
    constexpr void operator()(int, double) noexcept;
};

struct templated_function_call_operator {
    template <class... Args>
    constexpr void operator()(Args&&...) noexcept;
};

struct mixed_function_call_operator
: overloaded_function_call_operator
, templated_function_call_operator {
};

template <class T>
struct has_function_call_operator: std::false_type {};

template <class T>
requires std::is_member_function_pointer_v<decltype(&T::operator())>
struct has_function_call_operator<T>: std::true_type {};

template <class T>
inline constexpr bool has_function_call_operator_v 
= has_function_call_operator<T>::value;

int main(int argc, char* argv[]) {
    std::cout << has_function_call_operator_v<no_function_call_operator>;
    std::cout << has_function_call_operator_v<one_function_call_operator>;
    std::cout << has_function_call_operator_v<overloaded_function_call_operator>;
    std::cout << has_function_call_operator_v<templated_function_call_operator>;
    std::cout << has_function_call_operator_v<mixed_function_call_operator>;
    std::cout << std::endl;
}

目前它打印 01000 而不是 01111。如果它在最广泛的意义上是不可行的，那么可以假设 T 是可继承的（如果有帮助的话）。只要完全符合 C++20 标准，就可以使用最奇怪的模板元编程技巧。

Answer 1

&T::operator() 对于 3 个失败案例是不明确的。

所以你发现的特征是有一个明确的operator()

由于您允许 T 不是 final，我们可能会将您的特征应用到（假）class 以及现有的继承 operator() 和 class 来测试：

template <class T>
struct has_one_function_call_operator: std::false_type {};

template <class T>
requires std::is_member_function_pointer_v<decltype(&T::operator())>
struct has_one_function_call_operator<T>: std::true_type {};

struct WithOp
{
    void operator()() const;  
};

template <typename T>
struct Mixin : T, WithOp {};

// if T has no `operator()`, Mixin<T> has unambiguous `operator()` coming from `WithOp`
// else Mixin<T> has ambiguous `operator()`
template <class T>
using has_function_call_operator =
    std::bool_constant<!has_one_function_call_operator<Mixin<T>>::value>;

template <class T>
inline constexpr bool has_function_call_operator_v 
= has_function_call_operator<T>::value;

Demo