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

Question

C++ is mostly a superset of C, but not always. In particular, while enumeration values in both C and C++ implicitly convert into int, the reverse isn\'t true: only in C do i

Answer 1

Either leave the result as an int or static_cast:

Foo x = static_cast<Foo>(Foo_First | Foo_Second); // not an error in C++

Answer 2

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.

Answer 3

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.