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19/07/07 - 411 Views - No Ratings Yet

Level - Intermediate
      

Question -

What will be  the output of the following code?

#include <stdio.h>

int main()  
{    

  int a[][3] = { 1,2,3 ,4,5,6};    
  int (*ptr)[3] =a;      
  printf("%d\n\n %d "  ,(*ptr)[1], (*ptr)[2] );      
  ++ptr;    
  printf("\n\n%d\n\n%d"  ,(*ptr)[1], (*ptr)[2] );  
  return 0;
}  

Answer -

2


3


5


6

Explanation -

Well this output can be easily expained on applying the fundamentals of c programming.

In statement -

int (*ptr) [3] = a ;

We are just declaring a int array of size 3 whose base address is address of array 'a'. So ptr [0] will be equal to a[0] and ptr[1] will be equal to a[1] and finally a[2] will be equal to ptr[2], so in first printf statement we are able to produce output 2 and 3.

Now when following statement execute -

++ptr;

ptr is incremented by one and using normal pointer artihemetic it will jump by 3 integer size memmory locations so now ptr [0] will be pointing to a[3], ptr [1] to a[4] & ptr [2] to a[5].

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