Pointer arithmetic output asked in an interview

Question

char s[] = "arista2015";
char *p = s;
printf("%s",p+p[4]-p[1]);

This program gives the output as

ista2015

Can somebody explain the output?

You are aware that you wouldn't write code like that in a real program. — Jabberwocky, Jul 28 '15 at 14:17
p+p[4]-p[1] is nothing to write in the first place, that seems really wrong. What were you trying to do ? — Senua, Jul 28 '15 at 14:21
It was just a question asked in an interview, Since I could not answer it, I asked it here. — shivam mitra, Jul 28 '15 at 14:46
I really don't understand people down voting my question. It was just an interview question that I was not able to understand. — shivam mitra, Jul 28 '15 at 14:51
@shivammitra - because it's rubbish code with no lasting value for SO users. — Martin James, Jul 28 '15 at 15:49
The correct interview answer would be something like "I don't know or care. I never write code like that and I never will, I write code that I, and anyone who comes after me, can easily understand and maintain". — Martin James, Jul 28 '15 at 15:53

Lynn · Answer 1 · 2015-07-28T14:48:43.857

7

p[4] equals 't'. Its ASCII code is 116.

p[1] equals 'r'. Its ASCII code is 114.

Thus ~~p+p[4]-p[1] is p+2, i.e. 2 bytes past where p is pointing:~~ EDIT: Matt's answer brings up a very good point -- pointer arithmetic outside the string is undefined behavior, too, so p+116-114 and p+2 aren't actually guaranteed to be the same thing.

'a' 'r' 'i' 's' 't' 'a' '2' '0' '1' '5' '\0'
 ^       ^
 p      p+2

Amusingly, this is undefined behavior, though! On an EBCDIC system, it would print an empty string, as there 't' - 'r' == 10 (yes, really). The C99 standard only guarantees that the codes corresponding to the decimal digits '0', '1', '2'... '9' are consecutive.

edited Jul 28 '15 at 14:48

answered Jul 28 '15 at 14:10

Lynn

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I got it. But p[4] = 't' and p[1] = 'r'. So answer will be p+2. – shivam mitra Jul 28 '15 at 14:13
3

Correct -- under the assumption that the platform uses ASCII as execution character set. – undur_gongor Jul 28 '15 at 14:15

score 6 · Answer 2 · edited May 23 '17 at 12:23

6

Since the additive operators are left-right associative, the crucial expression is:

(p + p[4]) - p[1]

and not p + (p[4] - p[1]) as suggested by other answers/comments. Since p + p[4] is well outside the bounds of s, this causes undefined behaviour, which means that anything can happen (including, but not limited to, any particular output).

edited May 23 '17 at 12:23

Community

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answered Jul 28 '15 at 14:26

M.M

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Are you sure this causes undefined behaviour ? p + p[4] is never deferenced, p[1] is substracted from it before it is passed to printf (does it cause undefined behavior to have an invalid pointer as an intermediate result even if it's not deferenced ?) – BrunoLevy Jul 28 '15 at 14:44
1

@BrunoLevy Yes. The wording in the standard is "If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined.". C11, 6.5.6 This refers to adding an integer to a pointer. – Art Jul 28 '15 at 15:06

score 1 · Answer 3 · answered Jul 28 '15 at 15:08

Try to run these code and study the output

#include<stdio.h>
int main(){
char s[] = "arista2015";
char *p = s;
printf("Address of p: %d\n",p);
printf("Address of 'a' in arista2015 : %d\n",&s[0]);
printf("Address of'r' in arista2015 : %d\n",&s[1]);

p=p+p[4]-p[1]; // Now address of P becomes p + 2 so its points to address of 'i'
printf("Address of 'i....'%d\n",&s[2]);// and it print from address of i before null
printf("%d\n",p);//prints the address
printf("%s\n",p);//prints the value

 }

Run These code and check how its working as explain above..

my Output:
Address of p: 2686737
Address of 'a' in arista2015 : 2686737
Address of'r' in arista2015 : 2686738
Address of 'i....'2686739
2686739
ista2015

Pointer arithmetic output asked in an interview

3 Answers3