What is the difference between the dot (.) operator and -> in C++?

The dot and arrow operator are both used in C++ to access the members of a class. They are just used in different scenarios. In C++, types declared as class, struct, or union are considered "of class type". So the following refers to all three of them.

• a.b is only used if b is a member of the object (or reference[1] to an object) a. So for a.b, a will always be an actual object (or a reference to an object) of a class.
• a→b is essentially a shorthand notation for (*a).b, ie, if a is a pointer to an object, then a→b is accessing the property b of the object that a points to.

Note that . is not overloadable. → is overloadable operator, so we can define our own function(operator→()) that should be called when this operator is used. so if a is an object of a class that overloads operator→ (common such types are smart pointers and iterators), then the meaning is whatever the class designer implemented.

[1] References are, semantically, aliases to objects, so I should have added "or reference to a pointer" to the #3 as well. However, I thought this would be more confusing than helpful, since references to pointers (T*&) are rarely ever used.

example

#include<iostream>
class A {
   public: int b;
   A() { b = 5; }
};

int main() {
   A a = A();
   A* x = &a;
   std::cout << "a.b = " << a.b << "\n";
   std::cout << "x->b = " << x->b << "\n";
   return 0;
}

Output

This will give the output −

5
5

Smita Kapse

e

Updated on: 11-Feb-2020

2K+ Views

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