Double pointers - output explanation

Question

#include <stdio.h>
int main(void){

    int a=1,b=2,c=3; int *p,*q,**r; p=&a;
    r=&q;
    q=&c;
    a=*q+**r;
    printf("x=%d y=%d z=%d\n",**r,*p,*q); 
    *r=p;
    a=*q+**r;
    printf("w=%d\n",a);
    return 0;
}

Output:

x=3 y=6 z=3
w=12

I was able to predict the output correctly, but I am not sure whether I have the correct explanation for the output of z.

Please see whether I have the correct understanding:

• Right before *r=p; executes, we have a=6,b=2,c=3.
• When *r=p; executes, the value at the place to which r points to gets changed to p.
• Now r points to q that has address of c, so now q has address of a because p points to a. So q now points to a. So *q gives 6.
• Since r still points to q, and q points to a, **r gives 6.
• So *q + **r = 6+6=12

Is this the correct explanation?

Answer 1

Let's break it down:

1. int a=1,b=2,c=3;

   a = 1
   b = 2
   c = 3
   

2. int *p,*q,**r;

   p --> ?
   q --> ?
   r --> ? --> ?
   

3. p=&a;

   p --> a = 1
   q --> ?
   r --> ? --> ?
   

4. r=&q;

   p --> a = 1
   q --> ?
   r --> q --> ?
   

5. q=&c;

   p --> a = 1
   q --> c = 3
   r --> q --> c = 3
   

6. a=*q+**r;, as can be seen from the previous point, we have:

   a = 3 + 3 = 6
   

7. printf("x=%d y=%d z=%d\n",**r,*p,*q); prints:

   x=3 y=6 z=3
   

8. *r=p;. This is sneaky: since *r == q, this changes q. This is the same as doing q=p. So we get:

   p --> a = 6
   q --> a = 6
   r --> q --> a = 6
   

9. a=*q+**r;, as can be seen from the previous point, we have:

   a = 6 + 6 = 12
   

10. printf("w=%d\n",a); prints:

    w=12
    

Conclusion: your explanation is correct.

Answer 2

Yes. *r = p would make the item r points to, q, point to the same address as p, which is &a. **r is the same as *q which is the value at &a which is 6.