What is meant when you say you dereference a pointer variable?

Answer 1

When you dereference a pointer you are referring to the object being pointed at.

int* ptr = new int (42);

int& ref = *ptr;

You must be careful when you dereference pointers like this because ptr is volatile and could be deleted at any time. If ref is still in scope after ptr is released, ref will refer to invalid memory and thus invalidates your program. Moreover, if ptr falls from scope without releasing the memory it points to and ref remains in scope, there's no way to release the memory without invalidating ref which, again, invalidates your program. In order to be valid, references to dereferenced pointer values must be placed in a separate scope from the pointer and must fall from scope before the pointer falls from scope. Even if a reference falls from scope immediately after a pointer falls from scope, your program is still deemed invalid.

In some cases there is no need to hold a reference to a dereferenced pointer value, you can simply store the value by copying it to another variable.

int i = *ptr;

However, if the pointer is a complex object, keep in mind that this could invoke a cascade of copy constructors, which may impact performance.

The following code snippet shows all the major properties of a pointer: its own address, its value (the address being pointed to) and the indirect value (the dereferenced value).

std::cout << "Address: &ptr = 0x" << &ptr << std::endl;

std::cout << "Value: ptr = 0x" << ptr << std::endl;

std::cout << "Dereference: *ptr = " << *ptr << std::endl;