C++ different -> and "."

Question

Possible Duplicate:
what is the difference between (.) dot operator and (->) arrow in c++

I'm trying to learn c++, but what I don't understand is the different between "->" and "." when calling a method.

For example, I have seen something like class->method(), and class.method().

Thanks.

Time to get a [good book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) and find out. — GManNickG, Aug 11 '10 at 22:45

score 3 · Accepted Answer · answered Aug 11 '10 at 22:10

3

In a normal case, a->b is equivalent to (*a).b. If a is a pointer, -> dereferences it before accessing the element.

answered Aug 11 '10 at 22:10

viraptor

33,322
10
107
191

score 1 · Answer 2 · answered Aug 11 '10 at 22:10

1

The -> operator calls a method on the object pointed to by a pointer.

The . operator calls a method on an object itself.

If a is a pointer, a->b() is equivalent to (*a).b().

answered Aug 11 '10 at 22:10

SLaks

868,454
176
1,908
1,964

score 1 · Answer 3 · answered Aug 11 '10 at 22:10

1

You use -> when the thing on the left is a pointer to an object. You use '.' when the thing on the left is the object itself.

answered Aug 11 '10 at 22:10

NeilDurant

2,082
2
14
14

score 0 · Answer 4 · answered Aug 11 '10 at 22:10

0

When your calling a method through a pointer, you use ->; otherwise, use ..

Example:

MyClass* obj = new MyClass();
obj->myMethod();

MyClass obj2;
obj2.myMethod();

answered Aug 11 '10 at 22:10

Pedro d'Aquino

5,130
6
36
46

score 0 · Answer 5 · answered Aug 11 '10 at 22:11

It is maybe a bit irritating at first but C/C++ differenciates calling a method through the use of a pointer (use of ->) or through the use of a variable/reference (use of .). So the type of variable you use decides whether you call with the one or with the other option.

Note that in many other programming languages (Java, C#, ...) this is not the case.

score 0 · Answer 6 · answered Aug 11 '10 at 22:21

You will never see class.method() or class->method() in C++. What you will see is object.method(), obj_ptr->method() or class::method(). When you make an object of a class, you use the . operator, when refering to a pointer to an object, you use -> and when calling a static method directly without making an object, you use ::. If you have a class a_class like below, then:

class a_class
{
   public:
      void a_method()
      {
         std::cout<<"Hello world"<<std::endl;
      }
      static void a_static_method()
      {
         std::cout<<"Goodbye world"<<endl;
      }
}

int main()
{
    a_class a_object = a_class();
    a_class* a_pointer = new a_class();

    a_object.a_method(); //prints "Hello world"
    a_object->a_method(); //error
    a_object::a_method(); //error

    a_pointer.a_method(); //error
    a_pointer->a_method(); //prints "Hello world"
    a_pointer::a_method(); //error

    *a_pointer.a_method(); //prints "Hello world"
    *a_pointer->a_method(); //error
    *a_pointer::a_method(); //error

    a_class.a_method(); //error
    a_class->a_method(); //error
    a_class::a_method(); //error because a_method is not static        

    a_class.a_static_method(); //error
    a_class->a_static_method(); //error
    a_class::a_static_method(); //prints "Goodbye world"

}

C++ different -> and "."

6 Answers6