如何将枚举导入 C++ 中的不同命名空间?

我在命名空间中有一个枚举,我想像在不同的命名空间中一样使用它.直觉上,我认为我可以使用使用"或类型定义"来完成此操作,但实际上都不起作用.证明它的代码片段,在 GCC 和 Sun CC 上测试:

I have an enum in a namespace and I'd like to use it as if it were in a different namespace. Intuitively, I figured I could use 'using' or 'typedef' to accomplish this, but neither actually work. Code snippet to prove it, tested on GCC and Sun CC:

namespace foo
{

enum bar {
    A
};

}

namespace buzz
{
// Which of these two methods I use doesn't matter,
// the results are the same.
using foo::bar;
//typedef foo::bar bar;
}

int main()
{
    foo::bar f; // works
    foo::bar g = foo::A; // works

    buzz::bar x; // works
    //buzz::bar y = buzz::A; // doesn't work
    buzz::bar z = foo::A;
}

问题是枚举本身是导入的,但没有任何元素.不幸的是,在不破坏大量其他现有代码的情况下,我无法将原始枚举更改为包含在额外的虚拟命名空间或类中.我能想到的最佳解决方案是手动重现枚举:

The problem is that the enum itself is imported but none of its elements. Unfortunately, I can't change the original enum to be encased in an extra dummy namespace or class without breaking lots of other existing code. The best solution I can think of is to manually reproduce the enum:

namespace buzz
{
enum bar
{
    A = foo::A
};
}

但它违反了DRY 原则.有没有更好的办法?

But it violates the DRY principle. Is there a better way?

推荐答案

将现有命名空间包装在嵌套命名空间中,然后在原始命名空间中使用"该命名空间.

Wrap the existing namespace in a nested namespace which you then "use" in the original namespace.

namespace foo
{
    namespace bar_wrapper {
        enum bar {
            A
        };
    }
    using namespace bar_wrapper;
}

namespace buzz
{
    using namespace foo::bar_wrapper;
}

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