如何检测类中是否存在特定的PRIVATE成员变量？ [英] How to detect whether there is a specific PRIVATE member variable in class?

问题描述

此问题基于 。

目标：我想知道一个类是否具有成员变量 x 。我希望收到 true ，无论此变量是否是 private ， public 或受保护的。

Goal: I would like to know if a class has the member variable x. I would like to receive true regardless whether or not this variable is private, public or protected.

方法：如果某个班级有使用以下代码的成员变量：

Approach: You can get the information if a class has a member variable using the following code:

template <typename T, typename = int>
struct HasX : std::false_type { };

template <typename T>
struct HasX <T, decltype((void) T::x, 0)> : std::true_type { };

与constpr一起使用

Use it with

if constexpr (HasX<my_class>::value) {
   // do stuff with x
} else {
   // ...
}

上面的代码在这种情况下不起作用

The above code does not work in this case

struct my_class {
private:
   int x;
};

如何进行这项工作？我希望 HasX< my_class> :: value 为 true 。

How can I make this work? I would like HasX<my_class>::value to be true.

想法：

使用有权访问 T的 friend 类： ：x 。这似乎不起作用。查看此在线示例。

Use a friend class which has access to T::x. This does not seem to work. Check out this live example.

推荐答案

好吧...不确定此解决方案的正确性和局限性...但是...

Well... not sure about correctness and limits of this solution... but...

如果用<$定义了一个辅助结构c $ c> x 元素可访问

struct check_x_helper
 { int x; };

您可以编写从 check_x_helper 和您要查看的类是否包含 x 成员

you can write a template struct that inherit from both check_x_helper and the class you want to see if contain a x member

template <typename T>
struct check_x : public T, check_x_helper

内部 check_x 您可以如下声明（仅声明：在 decltype（）中使用）

Inside check_x you can declare (declare only: are used inside a decltype()) as follows

template <typename U = check_x, typename = decltype(U::x)>
static constexpr std::false_type check (int);

static constexpr std::true_type check (long);

观察第一个，即模板：当选中的类（ T ）包含 x 成员， decltype（U :: x）含糊不清，因为 x 继承自 T 和 check_x_helper ，因此该函数被SFINAE丢弃

Observe the first one, the template one: when the checked class (T) contains an x member, the decltype(U::x) is ambiguous because x is inherited from both T and check_x_helper, so this function is SFINAE discarded.

相反，当 T 不包含 x 成员时，没有任何歧义， decltype（U :: x）是 check_x_helper :: x 的类型（ int ）和第一个 check（）功能保持启用状态。

On contrary, when T doesn't contains an x member, there isn't an ambiguity, the decltype(U::x) is the type of check_x_helper::x (int) and the first check() function remain enabled.

现在，您需要使用type = decltype（check（0））;

Now you need something as

using type = decltype(check(0));

static constexpr auto value = type::value;

调用 check（0）（ int 参数表示对模板版本的偏好），并将检测到的值保存在静态constexpr

to call check(0) (the int parameter express the preference to the template version) and save the detected value in a static constexpr variable.

以下是完整的编译示例

#include <iostream>
#include <utility>

class foo
 { int x; };

struct bar
 { };

struct check_x_helper
 { int x; };

template <typename T>
struct check_x : public T, check_x_helper
 {
   template <typename U = check_x, typename = decltype(U::x)>
   static constexpr std::false_type check (int);

   static constexpr std::true_type check (long);

   using type = decltype(check(0));

   static constexpr auto value = type::value;
 };

int main()
 {
   std::cout << check_x<foo>::value << std::endl;
   std::cout << check_x<bar>::value << std::endl;
 }

此解决方案的缺点： decltype（U :: x） T 声明 x 作为方法或<$ c $时，code>也失败（含糊） c>使用类型。因此，给定

Drawback of this solution: decltype(U::x) fail (ambiguity) also when T declare x as a method or as a using type. So given

class foo
 { int x () { return 0;} ; };

或

class foo
 { using x = int; };

从 check_x< foo> :: value 中获得 1 。

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