CODEWITHSUNDEEP
c++enumsnested
wholerabbit
wholerabbit

Reputation: 11536

Access inner enum value outside namespace

Given the following minimal example:

foo.hpp:

class foo {       
    public:       
        enum bar {
            ONE,  
            TWO,  
            THREE 
        };        
        bar b;    
        foo ();   
};

foo.cpp:

#include "foo.hpp"       

foo::foo () : b(ONE) { }

How can I do what I'm trying to do below?

#include "foo.hpp"        

int main () {             
    foo *f = new foo();   
/* None of these work:    
    f->b = TWO;           
    f->b = foo::bar::TWO; 
    f->b = bar::TWO;      
*/                        
    return 0;             
}

I'm leaning toward the conclusion that this is not idiomatic in C++ and I must wrap the enum with the class in an outer namespace, or otherwise reorganize. What are the options and/or best practice?

Upvotes: 0

Views: 363

Answers (3)

Mark B
Mark B

Reputation: 96241

In C++ prior to C++11 enums (the enum name itself AND the values) are in the scope of the enclosing namespace/class.

So you would access it as: f->b = foo::TWO; (as seen in the other answer).

However you can utilize nested structs to make enum management a bit easier:

class foo {       
    public:       
        struct bar {
            enum Type {
            ONE,  
            TWO,  
            THREE  }
        };        
        bar::Type b;    
        foo ();   
};

Now you can qualify the name with your struct helper.

#include "foo.hpp"        

int main () {             
    foo *f = new foo();   
    f->b = foo::bar::TWO; 

    return 0;             
}

Upvotes: 1

Vlad from Moscow
Vlad from Moscow

Reputation: 310950

Enumerators are members of class. So you can write either as

f->b = foo::TWO;

or

f->b = f->TWO;

or even as

f->b = foo::bar::TWO;

provided that your compiler supports the C++ 2011 Standard

Upvotes: 0

marcinj
marcinj

Reputation: 49986

f->b = foo::TWO;

should work

Upvotes: 5

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