如何检查是否在C ++中声明了一个类？

问题描述

我正在围绕一个框架编写一个软件，并且我正在使用的类（准确地说，是扩展类）在以后的版本中已重命名。有什么方法可以在C ++ 11中编写一些宏/模板，以确定是否在代码中声明了具有特定名称的类？

I am writing a piece of software around a framework and a class I am using (to be precise, extending) was renamed in a later version. Is there some way to write some macros/templates in C++11 in order to determine if a class with a certain name had been declared at the point in code?

下面是我要完成的任务的说明。
假设文件class_include.h包含类 A 的定义：

An illustration of what I'm trying to accomplish follows.Let's say that file class_include.h contains either the definition of class A:

class A
{
...
};

或类 B ：

class B
{
...
};

和类 C 尝试扩展那些声明为：

and class C tries to extend whichever of those is declared:

#include <class_include.h>

#if (class A is declared)
class C : public A
#else // class B is declared
class C : public B
#endif
{
...
};

注意：我想尝试检查框架的版本，但是答案是问题使我感兴趣。我也不能更改任何框架头文件。

Note: It came to my mind to try and check a version of the framework but the answer to this question interests me. I also cannot change any framework header files.

编辑：可接受的答案取决于是否定义了类（这意味着声明），并且在我的情况下，如果定义了该类，则声明该类。

The accepted answer depends on whether the class is defined (which implies declaration) and, in my case, the class is declared iff it's defined.

推荐答案

可以，并且不需要宏。首先观察一下，即使可以使用完整的定义，也可以转发声明一个类。即这是有效的：

You can, and with no macros required. First an observation, you can "forward" declare a class even after its full definition is available. I.e. this is valid:

class foo{};
class foo;

现在，借助自制的 void_t 实现和 is_complete 类型的实用程序，您可以执行以下操作：

Now, with the help of a homebrew void_t implementation and an is_complete type utility, you can do something like this:

#include <type_traits>

template<typename... Ts> struct make_void { typedef void type;};
template<typename... Ts> using void_t = typename make_void<Ts...>::type;

template <typename T, typename Enabler = void>
struct is_complete : std::false_type {};

template <typename T>
struct is_complete<T, ::void_t<decltype(sizeof(T) != 0)>> : std::true_type {};

class A;
class B;

class C : public std::conditional<is_complete<A>::value, A, B>::type {
};

取决于 A 存在， C 将从 A 或 B 公开。 。

Depending on whether or not the full definition of A is present, C will inherit from A or B publicly. See a live example.

但是我告诫，这需要谨慎处理，否则您的程序中很可能会违反ODR。

But I caution, this needs to be handled with care or you are very likely to have an ODR-violation in your program.

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