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?

Marco Bonelli · Accepted Answer · 2019-10-30 23:09:43Z

2

Let's break it down:

int a=1,b=2,c=3;
```
a = 1
b = 2
c = 3
```
int *p,*q,**r;
```
p --> ?
q --> ?
r --> ? --> ?
```
p=&a;
```
p --> a = 1
q --> ?
r --> ? --> ?
```
r=&q;
```
p --> a = 1
q --> ?
r --> q --> ?
```

q=&c;

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

a=*q+**r;, as can be seen from the previous point, we have:
```
a = 3 + 3 = 6
```
printf("x=%d y=%d z=%d\n",**r,*p,*q); prints:
```
x=3 y=6 z=3
```
*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
```
a=*q+**r;, as can be seen from the previous point, we have:
```
a = 6 + 6 = 12
```
printf("w=%d\n",a); prints:
```
w=12
```

Conclusion: your explanation is correct.

edited Oct 30, 2019 at 23:09

answered Oct 30, 2019 at 23:03

Marco Bonelli

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Comments

Linus Gudmundsson · Accepted Answer · 2019-10-30 23:01:13Z

0

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.

answered Oct 30, 2019 at 23:01

Linus Gudmundsson

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