doubt resolving namespaces in enums?

Question

from the below question i sort of get how enums and namespace scoping works

Scope resolution operator on enums a compiler-specific extension?

However with regard to test code below i'm confused as to why in the below code snippet:

1) i can refer to return type in function signature as test_enum::foo_enum

2) however "using namespace test_enum::foo_enum" is not allowed

namespace  test_enum { 

   enum foo_enum { 

      INVALID, 
       VALID
    };
} 

// Case 1) this is allowed 
test_enum::foo_enum getvalue() {

     return test_enum::INVALID;

}

//Case 2) is not allowed 

using namespace test_enum::foo_enum;

is there a particular reason for not allowing case 2 ?
Also are "enums" more of C style construct and better to avoid in C++ code ?

Marlon · Accepted Answer · 2011-05-24 02:32:34Z

4

The reason using namespace test_enum::foo_enum; is not allowed is because foo_enum is not a namespace, it is an enum. What works is using test_enum::foo_enum;

I believe what you are trying to do is something like this:

namespace foo_enum {
    enum foo_enum_t {
        INVALID,
        VALID,
    };
}

using foo_enum::foo_enum_t;

This allows you to throw around foo_enum_t freely, but you still have to type out foo_enum::INVALID or foo_enum::VALID

answered May 24, 2011 at 2:32

Marlon

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2 Comments

bobbymcr Over a year ago

I agree with the namespace-scoped enum pattern. We have a similar coding guideline in place on my team.

Tony Delroy Over a year ago

POSIX reserves identifiers ending in _t, so they're best avoided in code intending to be portable.

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