实现可调用对象返回类型提取Trait时的SFINAE错误排查

背景 #

我正在练习SFINAE技术，尝试获取各类可调用对象（函数、函数指针、lambda、重载()运算符的对象）的相关信息，编写了如下代码：

#include <iostream>
#include <functional>
#include <string>
#include <type_traits>

template<typename Callable, typename = void>
struct function_type_info;

template<typename R, typename G, typename... T>
struct function_type_info<R(G::*)(T...)> {
    using return_type = R;
};

template<typename C>
struct function_type_info<C, typename std::enable_if<!std::is_member_function_pointer<C>::value, C>::type> {
    using return_type = typename function_type_info<decltype(&C::operator())>::return_type;
};

struct Callback {
    int operator()(double, std::string) {
        return 0;
    }
};

int main(int argc, char** argv) {
    auto b = [](double, std::string) -> int { return 0; };
    auto c = Callback{};

    std::cout << std::boolalpha;

    std::cout << std::is_same<function_type_info<decltype(&Callback::operator())>::return_type, int>::value << std::endl;
    std::cout << std::is_same<function_type_info<decltype(b)>::return_type, int>::value << std::endl;
    std::cout << std::is_same<function_type_info<decltype(c)>::return_type, int>::value << std::endl;

    return 0;
}

初始编译错误 #

编译器报出如下错误：

<source>:32:64: error: incomplete type 'function_type_info<main(int, char**)::<lambda(double, std::string)> >' used in nested name specifier
   32 |     std::cout << std::is_same<function_type_info<decltype(b)>::return_type, int>::value << std::endl;
      |                                                                ^~~~~~~~~~~
<source>:32:80: error: template argument 1 is invalid
   32 |     std::cout << std::is_same<function_type_info<decltype(b)>::return_type, int>::value << std::endl;
      |                                                                                ^
<source>:33:64: error: incomplete type 'function_type_info<Callback>' used in nested name specifier
   33 |     std::cout << std::is_same<function_type_info<decltype(c)>::return_type, int>::value << std::endl;
      |                                                                ^~~~~~~~~~~
<source>:33:80: error: template argument 1 is invalid
   33 |     std::cout << std::is_same<function_type_info<decltype(c)>::return_type, int>::value << std::endl;

调试后的问题 #

我无法理解错误原因：对于lambda和Callback对象，第二个function_type_info特化应该被匹配，因为std::is_member_function_pointer<C>::value为false，enable_if应生效。将enable_if的第二个参数改为void后，Callback可以正常工作，但lambda仍报错，错误信息如下：

<source>: In instantiation of 'struct function_type_info<main(int, char**)::<lambda(double, std::string)> >':
<source>:32:62:   required from here
<source>:16:11: error: invalid use of incomplete type 'struct function_type_info<int (main(int, char**)::<lambda(double, std::string)>::*)(double, std::__cxx11::basic_string<char>) const, void>'
   16 |     using return_type = typename function_type_info<decltype(&C::operator())>::return_type;
      |           ^~~~~~~~~~~
<source>:7:8: note: declaration of 'struct function_type_info<int (main(int, char**)::<lambda(double, std::string)>::*)(double, std::__cxx11::basic_string<char>) const, void>'
    7 | struct function_type_info;
      |        ^~~~~~~~~~~~~~~~~~
<source>: In function 'int main(int, char**)':
<source>:32:80: error: template argument 1 is invalid
   32 |     std::cout << std::is_same<function_type_info<decltype(b)>::return_type, int>::value << std::endl;

补充说明 #

仿佛无法访问lambda的()运算符。注：我知道可用std::result_of/std::invoke_result实现，但这是练习。

内容的提问来源于stack exchange，提问作者sunmat