有条件地启用静态成员，而不更改成员作用域

问题描述

我想为类启用静态成员，而不更改其范围。
考虑下面的抽象例子：

I want to enable a static member for a class without changing its scope.Consider the following abstracted example:

template<uint R, uint C>
class Foo
{
    static Foo ID;

    /* other members */
};

现在我想使用静态成员：

Now I want to use the static member like:

Foo<3, 3> foo = Foo<3, 3>::ID;

问题是 ID 仅当 R == C 时才存在。

（ Foo 实际上是 Matrix 和 ID 其唯一的方形矩阵存在）

The problem is that the ID field can only exist when R == C.
(Foo is actually a Matrix and ID its identity which only exists for square matrices)

我必须有条件地启用静态 ID 成员时满足条件。我现在的解决方案是这样：

So I have to conditionally enable the static ID member when the condition is met. My current solution is something like this:

struct EmptyBase { };

template<uint R, uint C>
class _Foo_Square
{
    static Foo<R, C> ID;
};

template<uint R, uint C>
class Foo : public std::conditional<R == C, _Foo_Square<R, C>, EmptyBase>::type
{
    /* other members */
};

但现在我不能写 Foo< 3，3& / code>来访问它。我必须写 _Foo_Square< 3，3> :: ID 。

不幸的是，我的应用程序的设计迫使它可以通过 Foo 类范围访问。
如果不是条件成员，我可以在Foo类中使用_Foo_Square< 3，3> :: ID; 来写。

Unfortunately the design of my application forces it to be accessible by the Foo class scope.If it wasn't a conditional member I could write using _Foo_Square<3, 3>::ID; in the Foo class.

这个问题有没有解决方案？

Is there a solution to this problem?

推荐答案

添加正确的前向声明并声明静态类成员为 public ，以下编译没有gcc 6.1.1的问题：

After adding a proper forward declaration, and declaring the static class member as public, the following compiles without issues with gcc 6.1.1:

#include <utility>

struct EmptyBase { };

template<int R, int C> class Foo;

template<int R, int C>
class _Foo_Square
{
public:

    static Foo<R, C> ID;
};

template<int R, int C>
class Foo : public std::conditional<R == C, _Foo_Square<R, C>, EmptyBase>::type
{
    /* other members */
};

void foobar()
{
    Foo<3, 3> a=Foo<3, 3>::ID;
}

这篇关于有条件地启用静态成员，而不更改成员作用域的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！