pointer c++ on objects and variables

Question

since I am currently learning pointers in C++, I am wondering about the syntax. So when I want to use a pointer on an object, I do not have to dereference it, in order to access the class attributes. When I just want to use a pointer on a simple variable, I have to use * for changing its values.

So why do I not have to use the * for the object? Because so I thought I would just change the memory address.

Object:

int age = 20;
User John(age);    
User *ptrUser = &John;

ptrUser->printAge(); // 20 (why no *ptrUser->printAge()  ??? )
cout << ptrUser // 0x123456...

Variables:

int a = 10;
int *ptrA = &a;   
*ptrA = 20;  
cout << a // 20

Thank you very much!

You might want to step back and [find a good book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). — crashmstr, Mar 10 '17 at 13:47
It seems to me that you're not noticing that "." and "->' are different operators. — Transcendental, Mar 10 '17 at 13:47

score 7 · Accepted Answer · answered Mar 10 '17 at 13:49

7

You have to dereference the pointer, the -> operator is just syntactic sugar for dereference-and-member-access:

Instead of

ptrUser->printAge();

you could write

(*ptrUser).printAge();

answered Mar 10 '17 at 13:49

alain

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score 0 · Answer 2 · edited May 23 '17 at 12:25

There can actually be two things:

You aren't aware about the -> operator:

As @alain pointed out in his answer, -> is mere syntactic sugar and can be replaced with a * and . as:

(*ptrUser).printAge();

You are aware about the -> operator and wondering why not *ptrUser->printAge():

Here, when you say *ptrUser = &John;, you are storing the address of John in a pointer variable called ptrUser. In other words, ptrUser points to John. So, when you want to access the member function printAge() of John, you can simply do:

ptrUser->printAge();

because ptrUser is already pointing to John.

Hope this is helpful.

pointer c++ on objects and variables

2 Answers2