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Home/ Questions/Q 772515
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T18:48:52+00:00 2026-05-14T18:48:52+00:00

In following code, #include<stdio.h> int main() { short a[2]={5,10}; short *p=&a[1]; short *dp=&p; printf(%p\n,p);

  • 0

In following code,

#include<stdio.h>   
int main()  
{  
  short a[2]={5,10};  
  short *p=&a[1];  
  short *dp=&p;  
  printf("%p\n",p);  
  printf("%p\n",p+1);  
  printf("%p\n",dp);  
  printf("%p\n",dp+1);  
}  

Now the output I got was :
0xbfb45e0a
0xbfb45e0c
0xbfb45e04
0xbfb45e06

Here I understood p and p+1, but when we do dp+1, then since dp points to pointer to short,
and since pointer to short is 4 bytes in size, so dp+1 should increase by 4 units but it
is increasing only by 2.
Please explain reason.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T18:48:52+00:00Added an answer on May 14, 2026 at 6:48 pm

    dp is defined as a pointer to a short and a short is two bytes. That’s all the compiler cares about. To actually make dp a pointer to a pointer to a short, you need to do

    short **dp = &p;
    
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