What's the difference between these two statements

Question

What's the difference between these two statements?

ob.A::ar[0] = 200;
ob.ar[0] = 200;

where ob is an object of class A

class A
{
    public:
        int *ar;
        A()
        {
            ar = new int[100];
        }
};

Answer 1

There is no difference. The explicit namespace qualification of ar is redundant in this case.

It might not be redundant in cases where (multiple, nonvirtual) inheritance redefines the name ar . Sample (contrived):

#include <string>

class A 
{
    public:
        int *ar;
        A() { ar = new int[100]; }
        // unrelated, but prevent leaks: (Rule Of Three)
       ~A() { delete[] ar; }
    private:
        A(A const&);
        A& operator=(A const&);
};

class B : public A
{
    public:
        std::string ar[12];
};


int main()
{
    B ob;
    ob.A::ar[0] = 200;
    ob.ar[0] = "hello world";
}

See it on http://liveworkspace.org/code/d25889333ec378e1382cb5af5ad7c203

Answer 2

In this case there is no difference.

This notation:

obj.Class::member

is only to solve the ambiguities coming from inheritance:

class A {
public:
  int a;
}

class B {
public:
  int a;
}

class C : public A, B {
   void func() {
      // this objects holds 2 instance variables of name "a" inherited from A and B
      this->A::a = 1;
      this->B::a = 2;
   }

}

Answer 3

In this case there is no difference. However imagine that ob is of class C that inherits both from class A and class B and both A and B have a field ar. Then there will be no other way to access ar but to explicitly specify which of the inherited data members you are refering to.

Answer 4

You are supposed to read this

ob.A::ar[0] = 200;
ob.ar[0] = 200;

like this

A::ar[0] it's the same thing for ar[0] , so the 2 rows are basically the same thing and the operator :: is used to indicate the so called resolution of the namespace , or just the namespace.

since ob is an object of type A, the namespace resolution is implicit and you do not need A:: before accessing ar[0] .