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Another easy one...

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:

g_pDD->Release();

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

Thanks,  

Raydot.
0
Raydot
Asked:
Raydot
1 Solution
 
jkrCommented:
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;

c.DoSomething();

is equivalent to

p->DoSomething();
0
 
Answers2000Commented:
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

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

raydot,

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...

Larry
0
 
RaydotAuthor Commented:
Wow, you guys need to lay off the coffee!  That was the fastest I ever got an answer on anything, my whole life.  Thanks...

Raydot.
0
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