Why does C++ need the scope resolution operator?

Question

(I know what the scope resolution operator does, and how and when to use it.)

Why does C++ have the :: operator, instead of using the . operator for this purpose? Java doesn't have a separate operator, and works fine. Is there some difference between C++ and Java that means C++ requires a separate operator in order to be parsable?

My only guess is that :: is needed for precedence reasons, but I can't think why it needs to have higher precedence than, say, .. The only situation I can think it would is so that something like

a.b::c;

would be parsed as

a.(b::c);

, but I can't think of any situation in which syntax like this would be legal anyway.

Maybe it's just a case of "they do different things, so they might as well look different". But that doesn't explain why :: has higher precedence than ..

question from:https://stackoverflow.com/questions/9338217/why-does-c-need-the-scope-resolution-operator

Answer 1

Why C++ doesn't use . where it uses ::, is because this is how the language is defined. One plausible reason could be, to refer to the global namespace using the syntax ::a as shown below:

int a = 10;
namespace M
{
    int a = 20;
    namespace N
    {
           int a = 30;
           void f()
           {
              int x = a; //a refers to the name inside N, same as M::N::a
              int y = M::a; //M::a refers to the name inside M
              int z = ::a; //::a refers to the name in the global namespace

              std::cout<< x <<","<< y <<","<< z <<std::endl; //30,20,10
           }
    }
}

Online Demo

I don't know how Java solves this. I don't even know if in Java there is global namespace. In C#, you refer to global name using the syntax global::a, which means even C# has :: operator.

---

but I can't think of any situation in which syntax like this would be legal anyway.

Who said syntax like a.b::c is not legal?

Consider these classes:

struct A
{
    void f() { std::cout << "A::f()" << std::endl; }
};

struct B : A
{
    void f(int) { std::cout << "B::f(int)" << std::endl; }
};

Now see this (ideone):

B b;
b.f(10); //ok
b.f();   //error - as the function is hidden

b.f() cannot be called like that, as the function is hidden, and the GCC gives this error message:

error: no matching function for call to ‘B::f()’

In order to call b.f() (or rather A::f()), you need scope resolution operator:

b.A::f(); //ok - explicitly selecting the hidden function using scope resolution

Demo at ideone