CODEWITHSUNDEEP
c++objectpointers
Ren
Ren

Reputation: 4683

Difference between treating the object as such or as pointer?

So say I have an object/pointer/whatever the definition of such a thing is:

A* a = new A();

who happens to have methods

b();
c();

The way of doing things that I've found out is this:

a->b();

and the method worked very well. However now I have seen people doing it this way:

(*a).b();

The question is: What is the difference (i.e. how are addresses and values managed in each) between these two ways of calling methods and according to that, which is best one to use?

If this a duplicate of other question, just let me know, I will erase it after I see the original question.

Upvotes: 1

Views: 136

Answers (4)

Robᵩ
Robᵩ

Reputation: 168626

For pointers, there is no difference:

If you declare:

A* pointer_to_A = new A();

Then these two are equivalent by definition:

pointer_to_A->b();
(*pointer_to_A).b();

If, however, you declare on object:

A a;

Then these two are not necessarily equivalent:

a->b();
(*a).b();

In this case, the first line invokes A::operator->() while the second invokes A::operator*(). (Aside: this case is somewhat rare. It is most often used for objects that behave like pointers: iterators, smart points and such. If they are well-designed, then the two forms above are still identical.)

Upvotes: 2

StackedCrooked
StackedCrooked

Reputation: 35485

The -> notation is syntactic sugar.

Upvotes: 0

Pubby
Pubby

Reputation: 53037

There is no difference. Prefer -> as it is cleaner and states what you mean better.

However, -> and * can be overloaded. So for certain classes they may do different things, although this is very very uncommon, and impossible for pointers.

Upvotes: 0

BYS2
BYS2

Reputation: 5259

There is no difference. It just a different notation.

Upvotes: 6

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