How should C bitflag enumerations be translated into C++?

时光毁灭记忆、已成空白 提交于 2019-12-01 09:23:29

Why not just cast the result back to a Foo?

Foo x = Foo(Foo_First | Foo_Second);

EDIT: I didn't understand the scope of your problem when I first answered this question. The above will work for doing a few spot fixes. For what you want to do, you will need to define a | operator that takes 2 Foo arguments and returns a Foo:

Foo operator|(Foo a, Foo b)
{
    return Foo(int(a) | int(b));
}

The int casts are there to prevent undesired recursion.

It sounds like an ideal application for a cast - it's up to you to tell the compiler that yes, you DO mean to instantiate a Foo with a random integer.

Of course, technically speaking, Foo_First | Foo_Second isn't a valid value for a Foo.

Either leave the result as an int or static_cast:

Foo x = static_cast<Foo>(Foo_First | Foo_Second); // not an error in C++
标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!