If an enum is declared in a function, what is the scope?

Question

Is the enum private to the function?

void doSomething()
{
   enum cmds {
      A, B, C
   }
}

Answer 1

The enum cmds type and the enumeration members A, B, and C, are local to the function scope.

Note that while enumerations don't introduce a new scope for their member constants, they do get scoped within the containing scope, in this case the function body.

int doSomething(void)
{
    enum cmds {
        A, B, C  // 0, 1, 2
    };

    return C;  // returns 2
}

int anotherThing(void)
{
    enum cmds {
        C, D, E  // 0, 1, 2
    };
    return C;  // returns 0
}

Run the example code here on Coliru

Answer 2

Like almost any other declaration, an enum type defined inside a block is local to that block.

In your case:

void doSomething()
{
   enum cmds {
      A, B, C
   }
}

the block happens to be the outermost block of the function, but it doesn't have to be:

void doSomething(void) {
    if (condition) {
        enum cmds {A, B, C};
        // The type and constants are visible here ...
    }
    // ... but not here.
}

(Goto labels are the only entity in C that actually has function scope.)