Another easy one...

Posted on 1998-10-06
Last Modified: 2010-04-01
What does the "->" operator do?  I keep coming across it, but it isn't covered in any of the books I've ever read on C++.  Here's a sample line from the DirectX SDK:


Does this put the variable g_pDD through the Release() function?


Question by:Raydot
LVL 86

Accepted Solution

jkr earned 50 total points
ID: 1174520
It accesses a class member function through a pointer to an instance of the class.

If you have e.g.
class C: public whatever
void DoSomething();

C c;
C* p = &c;


is equivalent to


Expert Comment

ID: 1174521
Damn! jkr types fast...

Guess I'll put this in as a comment since he beat me to submit button!

-> refers to the a member of a structure or class, when the structure/class is referenced by a pointer.

This is similar to the . operator used to refer to a member of struct/class.

-> is used when you have a pointer to a class/struct
 is used when you have a variable of the class/struct type

In your example g_pDD is a pointer to an object of some class

g_pDD->Release() calls the release member function of that object


Expert Comment

ID: 1174522
grf... guess i'm not the quickest draw in the east... (!)


this operator was inherited (sorry for the pun) from C; originally, it accessed a member of a struct through a pointer:

myStruct *foo;

foo->bar = 3;

of course, as jkr mentioned, in C++ it also allows you to access a member (data or function) of an instance of an object through a pointer to that instance...


Author Comment

ID: 1174523
Wow, you guys need to lay off the coffee!  That was the fastest I ever got an answer on anything, my whole life.  Thanks...


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