question for pointer

[mouse666666]

i just have a question are this the same if not why or why not.

x.y   // member y of object by x
x->y  menber y of object by x but am i passing by reference right?

[laserlight]

No, x->y is equivalent to (*x).y, i.e., member y of object pointed to by x.

[Swarvy]

When you use x.y, x is an object instance, but in the case x->y then x is a pointer to an object you have created and you are accessing member y through the pointer to that object rather than through the object instance itself.

x->y is the same as (*x).y
and
x.y is the same as (&x)->y

Edit: Laserlight is quick off the button when it comes to answering questions, lol.