如何包装std：：函数并轻松访问其返回和参数类型？

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

// 包装std::函数的模板类
template <typename Func>
struct FunctionWrapper;

// 特化模板类，用于包装函数指针
template <typename Ret, typename... Args>
struct FunctionWrapper<Ret(*)(Args...)> {
    using FunctionType = std::function<Ret(Args...)>;
    using ReturnType = Ret;
    using ArgumentTypes = std::tuple<Args...>;

    template <typename F>
    FunctionWrapper(F&& f) : func(std::forward<F>(f)) {}

    ReturnType operator()(Args... args) {
        return func(std::forward<Args>(args)...);
    }

    FunctionType func;
};

// 特化模板类，用于包装std::function
template <typename Ret, typename... Args>
struct FunctionWrapper<std::function<Ret(Args...)>> {
    using FunctionType = std::function<Ret(Args...)>;
    using ReturnType = Ret;
    using ArgumentTypes = std::tuple<Args...>;

    template <typename F>
    FunctionWrapper(F&& f) : func(std::forward<F>(f)) {}

    ReturnType operator()(Args... args) {
        return func(std::forward<Args>(args)...);
    }

    FunctionType func;
};

int add(int a, int b) {
    return a + b;
}

int main() {
    // 包装add函数
    FunctionWrapper<decltype(&add)> wrappedFunc(&add);

    // 访问返回类型
    using ReturnType = typename decltype(wrappedFunc)::ReturnType;
    std::cout << "Return type: " << typeid(ReturnType).name() << std::endl;

    // 访问参数类型
    using ArgumentTypes = typename decltype(wrappedFunc)::ArgumentTypes;
    std::cout << "Argument types: ";
    std::apply([](auto&&... args) {
        ((std::cout << typeid(args).name() << " "), ...);
    }, ArgumentTypes{}) << std::endl;

    // 调用包装的函数
    int result = wrappedFunc(3, 4);
    std::cout << "Result: " << result << std::endl;

    return 0;
}