c++ and injected base name

Question

The following code does not compile in gcc:

namespace One{
   class A{
   };
};

namespace Two{
   class A{
      public:
         void what(){
            cout << "Two::A says what!" << endl;
         }
   };

   class B : public One::A{
      public:
         B(){
            A xx;
            xx.what();
         }
   };

};

And it gives:

gccbug.cpp: In constructor ‘Two::B::B()’:
gccbug.cpp:23: error: ‘class One::A’ has no member named ‘what’

Now, I was told that this is correct behavior (due to injected base name of One::A making A refer to One::A). However, this code compiles in C# (well, after changing a few things), so this seems to be c++ specific.

What I'm wondering is.. why? Is there a specific purpose for injecting the base name "One::A" as "A"?

Just for reference, C# != C++. I'm not sure why you're comparing the two — Tony The Lion, Feb 26 '12 at 11:37
I can't say for sure (I don't have my copy with me right now), but as an aside, questions about why the C++ language was designed the way it was are often answered in a book called "The Design and Evolution of C++" by Stroustrup. — Stuart Golodetz, Feb 26 '12 at 12:07
@TonyTheLion, Well, it's the other c-like programming language that has "namespace", and one which I had access to a compiler to. I'm not saying there's only "One True Way" in which this is correct, but this did not seem intuitive, at least given this example. — kamziro, Feb 26 '12 at 12:28
@Tony The Lion Isn't the fact that C# != C++ what makes it _worth_ comparing them? — jogojapan, Feb 26 '12 at 12:41

Nawaz · Answer 1 · 2012-02-26T11:43:33.450

3

Is there a specific purpose for injecting the base name "One::A" as "A"?

Yes. It is so that you could write this:

namespace N
{
   class A
   {
       A *a;
   };
}

In the absence of injected-name, you've to write N::A *a which is not nice.

Note that it is because of injected-name, the following lines are allowed:

A::A *a1; //ok
A::A::A *a2; //ok 
A::A::A::A *a3; //ok 
A::A::A::A::A *a4; //ok 
//and so on

Online demo

edited Feb 26 '12 at 11:43

answered Feb 26 '12 at 11:38

Nawaz

353,942
115
666
851

Isn't this different from the situation explained by the asker? In the original example `class B` was inside one namespace, but the class it is derived from was in another. – jogojapan Feb 26 '12 at 11:51
1

@jogojapan But `class B` inherits all the names from its base class, including the injected name of the base class itself. Being in an inner scope, it hides names from the surrounding namespace. – Bo Persson Feb 26 '12 at 12:22
@Bo Persson Yes. I am just pointing out that the above example does not quite illustrate this point. In the above example the only reference to `class A` is inside the same namespace as the declaration of the class itself. You don't need name injection for that. – jogojapan Feb 26 '12 at 12:29
@jogojapan: You need name injection for that too. Otherwise, why does `A::A::A::A *a` work? – Nawaz Feb 26 '12 at 12:42
Ok, granted. There is name injection _at work_ there. The key difference I see between what I thought the question was about and the example above is that the question demonstrates that the compiler _prefers_ `One::A` over `Two::A`, rather than merely pointing out the name clash. What is the motivation for that preference? – jogojapan Feb 26 '12 at 13:36
@jogojapan: After my example code, there follows a natural conclusion : if `N::A *a` is not a nice syntax when declaring a member variable of `A` itself (means, the natural syntax should rather be `A *a` to convey the same meaning), then it becomes pretty much obvious why injected-name are preferred over any other name. I mean it would be absurd if the syntax `A *a` in the class `A` declares a pointer variable of type other than `A`. Don't you think so? – Nawaz Feb 26 '12 at 14:07
1

Well, given that there are two types called `A` in the original question, one is the base class but in a different namespace and the other is not the base class but in the same namespace, I still think it's a very ambiguous situation. But I am grateful (+1) for what I learnt through the discussion. – jogojapan Feb 27 '12 at 04:23

score 3 · Accepted Answer · answered Feb 26 '12 at 11:46

The only reason I can think of is that in C++ you are likely to refer to the base class name in the initializer list of the constructor, like this:

namespace Two {

  /*...*/

  class B : public One::A {
  public:
     B():A()
     {
        /*...*/
     }
   };
}

Of course the purpose then is different from the one in your example, because you actually declare a local variable inside the constructor, whereas in my example, the A() refers to the object of type A that is implicit in the definition of class B due to inheritance.

However, the situation of my example is more likely to occur, so I guess they thought let's not require the namespace to be made explicit in this case. As a consequence, any reference to A without a namespace is interpreted as referring to the base class, rather than any other class named A, even if it is in the same namespace as the declaration of B.

Ah, yes, that makes sense actually. Although I do say it feels a little like a "hack". — kamziro, Feb 26 '12 at 12:35

score 0 · Answer 3 · answered Feb 26 '12 at 11:38

0

By qualifying A with One:: you added the A from namespace one in scope, so the compiler will look there for it's name resolution.

answered Feb 26 '12 at 11:38

Tony The Lion

61,704
67
242
415

But the object of type `A` declared inside the constructor has not been qualified with `One::`. It is interpreted as belonging to `namespace One`, although the declaration of `class B` itself is actually inside `namespace Two`. – jogojapan Feb 26 '12 at 11:48

c++ and injected base name

3 Answers3