Pointers: a query about pointers

Question

I'm learning C and C#. I'm learning about pointers and don't know what it means to combine the indirection operator and the address operator. What does it mean to combine the two?

Here is an example:

    int *p, *q;
    p = *&q;

Answer 1

& can be thought of as an address of (<something>) operator. So &q is address of q. Now * can be thought of as an value at (<something>) operator. So *q is basically the value stored at the address contained in q, i.e, * treats the variable as always containing an address. Now *&q, by associativity is *(&q). Which means

value stored at (address of q) which is same as value stored at q

address of q will be having another address since q is a pointer. So it is the same as

p=q

Answer 2

It means what it must mean. :)

If you read it from left to right, the right hand side simply means "the value retreived by following the pointer whose value is the address of q".

It's the same as p = *(&q);, and thus it's the same as p = q;.

I didn't even notice, but your program is wrong to declare p and q to be pointers. That won't compile, it should be:

int p, q;

p = *&q;

Also, this might be a bit ill-defined since q is never assigned a value before being read, not 100% sure about that.

Answer 3

&q;

would give you the address of the variable q, that is it's a pointer to q we could write

int ** pointerToQ = &q;

if we then say

*pointerToQ

we are asking for whater ever pointerToQ points to, which is q itself. So

*&q 

Just gets us back to q.

Answer 4

Your example would set pointer p to the value of pointer q. It does nothing to the values pointed by p and q. It just sets the variable p, to the value of variable q, which both happen to be pointers to integer values. Lets see some examples:

*p = *q; // the value at the address pointed by q, gets copied to the address, pointed by p
p = &q; // this is illegal, since the type of &q is int ** (pointer to a pointer to an integer), but the type of p is int *