When you "de-pointer" a pointer to access it as if it were an object using the * operator right before the object's name, what exactly is it doing?
I ask this because I have pointers to objects that store a lot of data and I don't want C++ to copy it or do anything expensive. There's a fine line between me being able to copy this object and using the functions in it.
You get an lvalue referring to the object located at the address given by the pointer.
In general, dereferencing a pointer alone will never cause a copy. A copy occurs when the reference obtained is used to construct a new object of the same type (which happens implicitly when passing by value to a function), or used as the argument to an assignment operator.
when you dereference a pointer you are just accessing its element; you are not copying anything. These two calls do the same thing:
(*p).element = 1;
p->element = 1;
if you want a thorough explanation; you might take a look at this question
The dereference operator or indirection operator, denoted by "*" (i.e. an asterisk), is a unary operator found in C-like languages that include pointer variables. It operates on a pointer variable, and returns an l-value equivalent to the value at the pointer address. This is called "dereferencing" the pointer.
(http://en.wikipedia.org/wiki/Dereference_operator)
When you use pointers to point an object the pointers just point to the address of the object but it doesn't copy anything.....
'*' returns the object as an l-value this pointer is pointing. It's as simple as that.
