What value does a Pointer to a Pointer print if not a single dereference operator is used at all?

Question

I have tried the output of the following code. But, i don't yet understand what might the value of q represent.

I understand that the *q points to p implying that, printing *q would print the address of p, where as **q would print the value at the address p points to, i.e., x(=5).

#include <iostream>
using namespace std;
int main()
{
    int x=5;
    int *p,**q;
    p=&x;
    q=&p;
    cout<<q;
    return 0;
}

So what does q alone represent? What is that value that is printed when just q is printed?

Answer 1

You say:

printing *q would print the address of p

However this is not true. Printing q would print the address of p. After all, you assigned that in doing q = &p;.

Printing *q would print the value of p , which is the address of x.

Answer 2

Since q is a pointer to p, the value of q is the location of p, whatever that means on this particular platform. (Most likely the memory address containing p's value.)

Answer 3

(I will assume 64-bit pointers and 32-bit integers in this answer, just to have concrete values to write about. It holds with any vaues, though).

Let's analyse p first. p is 8 bytes storing an address (of the variable x). Printing *p will print the value residing at that address, which is the 4 bytes of the value of variable x. Printing just p will print the 8 bytes of the address stored in p.

Now, the same applies to q. q is 8 bytes storing an address of the variable p. So printing *p will print the value residing at that address, which is the 8 bytes of the value of p. Printing just q will print the 8 bytes of the address stored in q.

Answer 4

Just print value and ardess of your variable and you will see:

   x: 5                     &x: 0x7fff691dfcc4
   p: 0x7fff691dfcc4        &p: 0x7fff691dfcb8
   q: 0x7fff691dfcb8        &q: 0x7fff691dfcb0

&var - location; 
 var - value

Answer 5

int x = 1;

A block of memory is allocated to hold int value. And that block is named as 'x'. Let's say x is allocated at 0x1234.

int *p;

Here 'p' is pointer to int that means p would contain the address of some int. Let's say p is allocated address 0x2345.

p = &x;

This would make p contain the address of x i.e 0x1234 would be stored in the location allotted to p.

int **q;

Here q is pointer to pointer to an int that means q would contain the address of pointer to int. Let's say q is allotted address 0x3456.

q = &p;

This would make q contain the address of p i.e 0x2345 would be stored in the location allotted to q.

Hope I am simple and clear...

Answer 6

Pointer is just a data type, being not very different from other types. It stores an address of another data object, and you can use *p to access the data object. So, pointers are not different from other types but their contents have some special meaning.

(int**) is such a pointer that points to a data object of type (int*), being not very different from other pointers (they may point to data object of type int).

So, the content of q is a pointer to p. *q indicates you use the content to find p. the content of p is a pointer to an Integer, *p indicates you use the content to find the integer.

Answer 7

By adding ** before q in declaration of q, you are saying to compiler that, q is a pointer which will point to address of a pointer. So when you are printing q, you are actually printing address of p pointer, not the value of address p point to (i.e. x).

In your code, p is a pointer which is storing an address of an integer. But p itself has an address, which you are storing in q.