在 C++ 中组合数据成员访问

Question

我试图通过指向数据成员的指针迭代对象来访问数据成员。

想法是有一个可变参数模板函数，它在第一个 obj 上调用 std::invoke 并将结果传递给下一个指向数据成员的指针。就像是

compose x fnList = foldl obj (\x f -> f x) fnList

来自函数世界。

我一路上得到的东西是：

// main.hpp
#include <iostream>
#include <functional>
// initial template to stop recursion
template<typename T>
T getMember(T obj) {
  return obj;
}
// variadic template, where recursive application of 
// pointer to data member should happen
// I think return type should be something like "*Ret"
template<typename T, typename K, typename Ret, typename ... Args>
Ret getMember(T obj, K memberPointer, Args ... args) {
    return getMember(std::invoke(memberPointer, obj), args ...);
}

和

//main.cpp
#include <iostream>
#include "main.hpp"

//inner class
class Engine 
{
    public:
    std::string name;
};
// outer class
class Car
{
    public:
    int speed;
    Engine eng;
};

void main()
{
    Car car;
    car.speed = 1;        
    car.eng.name = "Some Engine Name";

    // should be same as call to id function, returning the only argument
    Car id = getMember(c1);
    // should "apply" pointer to data member to the object and
    // return speed of the car
    int speedOfCar = getMember(car, &Car::speed);
    // should "apply" pointer to data member to the car,
    // pass the resulting Engine further to &Engine::name,
    // return "Some Engine Name"
    std::string nameOfEngineOfCar = getMember(car, &Car::eng, &Engine::name);

    std::cout << nameOfEngineOfCar << std::endl;
}

编译无法推导出 return 类型 Ret（gcc 5+，C++17）有可能吗？有什么限制（可以在C++14中完成）吗？

Answer 1

您不能推导出只出现在函数 return 类型位置的模板参数。但是模板参数推导并不是 C++ 中唯一的一种推导。你可以这样做：

template <class Bar, class Baz>
auto foo(Bar x, Baz y) 
      -> decltype(auto) { 
    return moo(x, y); 
}

如果由于旧编译器而失败，更安全的回退是

template <class Bar, class Baz>
auto foo(Bar x, Baz y)
     -> decltype(moo(x, y)) { 
    return moo(x, y); 
}

在 C++ 中组合数据成员访问

Compose data member access in C++

c++

pointer-to-member

variadic-templates