Pointer to pointer issue in C

Question

This is what I'm trying to achieve. Inside pointer 'r' I have the address of pointer 'p', now I try to move the value of 'p'(which is an address) using pointer arithmetic by calling function moveforward(). I get this error :

void value not ignored as it ought to be
movingresult=moveforward(r);

What's wrong here? It's a bit complicated though dealing with pointer to pointer.

#include <stdio.h>

void moveforward(int** y) {
     *y = *y+1;
     printf("value of *y in function : %p\n",*y);

}


int main () {
int a = 16;
int *p;
int **r;
p = &a;
r = &p;
int **movingresult;

printf("Value of p before moving :%p\n",p);
movingresult=moveforward(r);
printf("Value of p after moving :%p",movingresult);

return 0;

}

I just changed my code , so now it looks like this, everything runs well , but the result is not what i expected.

#include <stdio.h>

void moveforward(int** y) {
 *y = *y+1;
 printf("Value of *y now has been moved to : %p\n",*y);

}


int main () {
int a = 16;
int *p;
int **r;
p = &a;
r = &p;
int **movingresult;

printf("Value of p before moving :%p\n",p);
moveforward(r);
printf("Value of p after moving :%p\n",r);

return 0;

}

OUTPUT :
Value of p before moving :0x7fffa5a1c184
Value of *y now after moving : 0x7fffa5a1c188
Value of p after moving :0x7fffa5a1c178

MY EXPECTATION : the 'p' after moving must be equal to *y which is 0x7fffa5a1c188

I'm getting this error when I compile it. main.c: In function 'main': main.c:19:13: error: void value not ignored as it ought to be movingresult=moveforward(r); ^ exit status 1 — Manav Dubey
– Manav Dubey, Commented Mar 22, 2017 at 1:40
printf("Address of p after moving :%p\n",r); --> printf("Address of p after moving :%p\n", p); or printf("Address of p after moving :%p\n",*r); — BLUEPIXY
– BLUEPIXY, Commented Mar 22, 2017 at 2:11
Why are you talking about "address of p" changing? You mean, the value of p (which is an address). — philipxy
– philipxy, Commented Mar 22, 2017 at 2:14

alex lee · Accepted Answer · 2017-03-22 03:49:27Z

2

when you call the function moveforward, the pointer address of r not change. moveforward(r); after call moveforward, the address of r isnot change yet. so if you execute printf("Value of p after moving :%p\n",r);

the address of r are still not change.

if you want get your expected value, you should do: printf("Value of p after moving :%p\n",*r);

answered Mar 22, 2017 at 3:49

alex lee

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Srinidhi Shankar · Accepted Answer · 2017-03-22 04:02:58Z

1

This is how things look before doing any pointer arithmetic.

After you execute moveforward(r);, this is how things look:

and these are the pointer values:

printf("Address pointed by p :%p\n",p); // Prints 0x7fffa5a1c188
printf("Address pointed by p :%p\n",*r); // Prints 0x7fffa5a1c188
printf("Address pointed by r :%p\n",r); // Prints 0x7fffa5a1c178

This is because, when you execute the following piece of code,

void moveforward(int** y) {
*y = *y+1;
printf("Value of *y now has been moved to : %p\n",*y);
}

Only the content of pointer p is incremented(0x7fffa5a1c184 to 0x7fffa5a1c188). But the content of pointer r (0x7fffa5a1c178) remains the same.

edited Mar 22, 2017 at 4:02

answered Mar 22, 2017 at 3:55

Srinidhi Shankar

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