shoe_copy=shoe1; will call the assignment operator, not the copy constructor. The copy constructor is called for example when you pass an object to a function by value.

// copy constructor called (notice I pass by value, not by reference)
void f(Shoe s)
{
}

yes but what willl shoe_copy=shoe1 do?

Call the copy constructor.

Let me explain:

Shoe shoe1; // calls the constructor
Shoe shoe_copy(shoe1); // copy constructor
Shoe shoe_copy = shoe1; // copy constructor, even if it seems it's the assignment operator;

shoe_copy = shoe1; // assignment operator

but the last example shoe_copy=shoe1; - does that set the adresses equal or what does it physically do to shoe_copy.

It will not modify and addresses but it will set all of the values of shoe_copy to the values of shoe1. It wont call a copy constructor. I have seen this called a "shallow copy" (here)

Ohhhh... okay that is probably the most informative clear article I have read on the subject and I thank you for that excellent link.

So when p=q calls the copy constructor is this accessing a compiler-defined ovverloaded version of the = operator?

sorry that last post was badly stated as a question, what I meant was: When p=q calls an overloaded assignment operator and preforms a shallow copy is this done using the automatic copy constructor that is called when an object is initialized to another?
So if i explicitly defined a complex copy constructor and said p=q that used the assignment operator would it preform a complex copy?

No, the copy constructor is used only when called explicitly or when an object is being initialized. You could, however, have the assignment operator call your copy constructor.

Yet another context where copy constructor works: