GATE CSE 2011 | Question: 22

Question

What does the following fragment of C program print?

    char c[] = "GATE2011";
    char *p = c;
    printf("%s", p + p[3] - p[1]);

• A. $\text{GATE2011}$
• B. $\text{E2011}$
• C. $2011$
• D. $011$

Comments

@arpit.jha yeah one extraa $*$ i used thnx for corrcting me

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$\text{ concept : $\underline{\text{pointer arithmetic}}$}$

char c [] = "GATE2011";
// indices:  01234567
char *p =c;

p+ p[3] - p[1] 

= 0 + E - A 

= 0+ 69 - 65

= 4 

 printf() prints from index 4 to end of the string so,the output is 2011. So, Option C)2011 is Correct.

reference : https://stackoverflow.com/questions/4989416/a-c-program-from-gate-paper

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This in indeed correct explanation but the 4 is appended to "p" hence taking "p" as zero is kinda  misdirection I think

 

p + E - A 

p + 4

so printf statement prints everything from the (p + 4)th locaiton till /0

Answer 1

2011 is the answer. 

In C, there is a rule that whatever character code be used by the compiler, codes of all alphabets and digits must be in order. So, if character code of '$A$' is $x$, then for '$B$' it must be $x+1$. 

Now $\%s$ means printf takes and address and prints all bytes starting from that address as characters till any byte becomes the code for $'\0'.$ Now, the passed value to printf here is
$p + p[3] - p[1]$

$p$ is the starting address of array $c. p[3] = 'E'$ and $p[1] = 'A'$. So, $p[3] - p[1] = 4$, and $p + 4$ will be pointing to the fifth position in the array c. So, printf starts printing from $2$ and prints $2011$. 

(Here $``\text{GATE2011}”$ is a string literal and by default a $'\0'$ is added at the end of it by the compiler). 

NB: In this question $\%s$ is not required. 

 printf(p + p[3] - p[1]);

Also gives the same result as first argument to printf is a character pointer and only if we want to pass more arguments we need to use a format string. 

Comments

everyone facing issues, this comment might help u alot

@utpal podder
"Bikram sir why are we considering the array index starts with 0 here why not 1 here...if we consider the array index starts with 1 here then what could be the ans then..." - it seems to be valid doubt
lets start the array indexing by 1
means p[3] - p[1] = ascii value of T - ascii value of G = 13
our array is of length 9 including '/0'
so p + 13 is out of the array index so nothing will print, which is not in options so indexing starts with 0
if the question had an option like empty string then it both options might be correct

@shenoy
"Why my thinking here is wrong? Why we are finding length of the pointer first by subtracting and then adding to the base address?"
operations which are allowed on pointers : 
1) inc/dec of a pointer
2) addition of integer to a pointer
3) subtraction of integer to a pointer
4) subtracting two pointer of same type
5) comparisons of pointer of the same type

so u can clearly see there is no such operations "addition of pointer to pointer"
and its very valid that why it isn't allowed, take a example by yourself u will get it

main doubt - why we are doing ascii code subtraction
"For pointer subtraction to happen, we need two pointers pointing to elements of the same array"
but here p[3] is pointing to character and same with p[1]
so we did ascii code subtraction

variation - 
if the question had
char *p1 = &c[3];
char *p2 = &c[1];
int diff = p1 - p2;
then we will do 3 - 1 = 2
 

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We don't actually need to know the ASCII values of 'A' and 'E'.

To calculate: 'E' - 'A'

Use the fact that the relative positioning between 'A' and 'E' doesn't change (or the distance between them remains the same).

Character	A	B	C	D	E
Index	1	2	3	4	5

 

Then we simply calculate \[ 5 - 1 = 4 \]

But of course, we are able to use this method because the english alphabet follows the usual sequence in ASCII too.

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how is p[3] - p[1] is 4 ?

Answer 2

Option c is right.

Answer 3

BEST ANSWER

Answer 4

char c[] ="GATE2011";

char *p = c;

printf("%s", p + p[3] - p[1]);

Concept - Pointer arithmetic 

p[3] = *(p+3) = E

p[1] = *(p+1) = A 

ASCII VALUE DIFFERENCE OF E-A = 4

so finally printf("%s", p +4]); here p is the pointing to the address  first element of char array c[ ] and p +4 will point to Address of P[4] element of array , printf takes initial address and prints untill "\0" is found, so final output will be  2011

Answer 5

Correction option: C