检查类是否有可能重载的函数调用运算符

Question

我想知道是否有可能在C++20中实现特征以检查类型T是否具有可能重载/可能是模板化的函数调用运算符：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;

这样，如下所示的代码将导致正确的结果：

#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;
}

当前，它打印01000，而不是01111。如果从最广泛的意义上说它不可行，则可以认为T是可继承的（如果有帮助的话）。只要它们完全符合C++20标准，就欢迎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。

所以您发现的特征是有明确的operator()

由于您允许T不为final，我们可能会将您的特征应用于具有现有继承的operator()的（假）类和要测试的类：

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