C code question related pointer

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C

int main(){
int  a[5];
int *ptr;
ptr=a;
for(int i=0;i<5;i++){
scanf("%d",ptr);
printf("%p\n",&ptr+i);
ptr++;}}

Output:

0x7fff1e80ea58
0x7fff1e80ea60
0x7fff1e80ea68

but if am trying to put

C

printf("%p",&ptr);

then i am getting the output as
Output:

0x7fff1e80ea58
0x7fff1e80ea58
0x7fff1e80ea58

Why ?

What I have tried:

I hav tried to replace ptr with a

Posted 27-May-22 18:41pm

sundresjh

Updated 27-May-22 22:23pm

OriginalGriff

v2

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OriginalGriff · Answer 1 · 2022-05-27T19:17:00

First off, always indent and space your code, even in a little scrap of program like this - it makes it a lot easier to read:

C

int main()
    {
    int  a[5];
    int *ptr;
    ptr = a;
    for(int i = 0;i < 5; i++)
        {
        scanf("%d", ptr);
        printf("%p\n", &ptr + i);
        ptr++;
        }
    }

The difference between

C

printf("%p\n", &ptr + i);

and

C

printf("%p\n", &ptr);

is obvious: in one you add a variable value i to static value &ptrand in the other you don't.

Think about it: ptr is a variable which contains the address of an array element in memory - but as a variable it is also stored somewhere, so it has it's own address &ptr which doesn't change when you alter the value the variable contains.

Try this:

C

int main()
    {
    int  a[5];
    int *ptr;
    ptr = a;
    printf("%p\n", ptr);
    printf("%p\n", a);
    printf("%p\n", &ptr);
    printf("%p\n", &a);
    }

And see what you get.

CPallini · Answer 2 · 2022-05-27T22:23:00

Solution 2

Try

C

#include <stdio.h>
  
int main()
{
  int  a[5];
  int *ptr = a;
  for(int i=0; i<5; i++)
  {
    scanf("%d",ptr);
    printf("address of pointer %p, address of pointed value %p, value %d\n", &ptr, ptr, *ptr);
    ptr++;
  }
}

Posted 27-May-22 22:23pm

CPallini