C programming

Question

include <stdio.h>

int main()
{
    char *c[]={"abhi","abhishek","abhisheksing"};
    
    printf("%s",(*(*(c+2)+7)));

    return 0;
}
 

What is the output?

Comments

We are given an array of pointers which points to the 3 strings.

abhi

abhishek

abhisheksing

Initially c points to the first cell. *(c+2) means that c now points to the 3rd cell.

The pointer is currently at 'a'.

a	b	h	i	s	h	e	k	s	i	n	g
0	1	2	3	4	5	6	7	8	9	10	11

Now when we add 7 to it it points to k and ksing gets printed.

The output is :ksing.

 

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No the output is Segmentation fault

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Yup right.. that extra * in " (*(*(c+2)+7)) " won't give the result as  (*(*(c+2)+7))  means 'k'. But using %s requires an address value and that is returned by  (*(c+2)+7). (*(*(c+2)+7))  returns a character and not an address. So printf("%c", (*(*(c+2)+7)) ) will work and not printf("%s", (*(*(c+2)+7)) ).

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yup  there is an extra *.. If %c is there the ans will be only k.

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Yes minipanda correct.

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@minipanda

1.Pleasesuggest if it should print with '%s' then what should be the correct statement for output ''ksing".

2.%c--> requires value at an address and it prints only charachar.

 %s-->requires starting address or any address of an array and it prints tills null character

 

3. (*(C+2)+7)---->&A[2][7]. Let A be an array.

 

Please suggest on above three points whether my understanding is correct?

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1. printf("%s", (*(c+2)+7) );  this will print ksing

2. Yes..both are right

3. Yes true..both printf("%s", (*(c+2)+7) ); AND printf("%s", &c[2][7] );  will give "ksing" and they mean the same.