Calling a static method by repeating the object name?

Question

I have a singleton:

struct foo {
  static foo& instance() {
    static foo f;
    return f;
  }
};

When re-arranging some code I ended

Answer 1

As stated in the other answers, the reason is name injection. To me, the main use case would be the following

struct B1 { void f(){} };
struct B2 { void f(){} };

struct D : B1, B2 { }

int main() {
    D obj; 
    obj.f(); 
}

In main the call to f is ambiguous and won't compile. The way to be specific is a qualified call, ie

obj.B1::f(); 

Answer 2

Because of [class]/2:

A class-name is inserted into the scope in which it is declared immediately after the class-name is seen. The class-name is also inserted into the scope of the class itself; this is known as the injected-class-name.

So foo::foo is an injected class name, denoting foo itself.

---

Actually it's a bit more complicated: according to [class.qual]/2, foo::foo alone denotes a constructor of foo. In order to denote a class, it should either be preceded by struct (making it elaborated-type-specifier), or followed by :: (making it a nested-name-specifier - this is your case), or be a base-specifier (for example struct bar : foo::foo {};).

Answer 3

It is due to "injected-name" — which means if foo is a class-name, and the same name "foo" is also injected into the class-scope which is why your code works. It is 100% Standard-conformant.

Here is one interesting example which shows the benefits of this feature:

namespace N
{
   //define a class here
   struct A 
   { 
       void f() { std::cout << "N::A" << std::endl; }
   };
}

namespace M
{
   //define another class with same name!
   struct A 
   { 
       void f() { std::cout << "M::A" << std::endl; }
   };

   struct B : N::A  //NOTE : deriving from N::A
   {
         B()
         {
            A a;
            a.f(); //what should it print?
         }
   };
}

What should a.f() call? What is the type of a? Is it M::A or N::A? The answer is, N::A, not M::A.

• Online Demo

It is because of name-injection, N::A is available inside the constructor of B without qualification. It also hides M::A, which remains outside the scope of B. If you want to use M::A, then you've to write M::A (or better ::M::A).