CODEWITHSUNDEEP
c
debug
debug

Reputation: 31

Why this c program output like this?

#include<stdio.h>
main
{
    int x[]={1,2,3,4,5};
    int i,*j;
    j=x;
    for(i=0;i<=4;i++)
    {
        printf("%u",j);
        j++;
    }
}

output:

65512
65514
65516
65518
65520

But when I change the printf to"

printf("%u",&j[i]);

Output is:

65512
65516
65520
65524
65528

Why the address differ by 2 in first case and 4 in second casee?

What is wrong with just printing j and printing &j[i]?

Upvotes: 1

Views: 126

Answers (2)

Igor
Igor

Reputation: 27268

First, just to make it clear, you are printing the pointer j, and not the pointed value, *j

Now, regarding the printed address. In your second example:

for(i=0;i<=4;i++)
{
  printf("%u",&j[i]); 
  j++;

&j[i] equals to (j+i). i is incremented in each iteration, which contributes 2 to the pointer's value, and j is incremented too, which contributes another 2.

Upvotes: 0

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272687

You get jumps of 4 in the second example because you are incrementing j and offsetting by i! Both of these contribute a difference of 2.

Note also that printf is not type-safe; it is up to you to ensure that the arguments match the format-specifiers. You have specified %u, but you've given it an int *, you should use %p for pointers.

Upvotes: 3

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