#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++;

Categories >> Software >> Programming Languages >> C
Suggest New Category

#include<stdio.h>
int main()
{
int i=2;
int j=++i + ++i + i++;
printf("%d\n",i);
printf("%d\n",j);
}

Question Posted / gaurav

15 Answers
30424 Views
Infosys, I also Faced
E-Mail Answers

Answers were Sorted based on User's Feedback

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / gaurav

5
12

Is This Answer Correct ?

35 Yes

14 No

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / mayank kumar

i=5
j=11

Is This Answer Correct ?

22 Yes

10 No

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / tamil selvam

i value is 2.
j value is 7.

Is This Answer Correct ?

19 Yes

9 No

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / gaurav

Please use gcc compiler....u will get answer 1,i.e. 5 and 12

Is This Answer Correct ?

10 Yes

3 No

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / rajesh kumar mahto

i=5 and j=11 (100% correct answer)
this is because
j=5+4+2(right to left execution)
i=5 due to latest updation of i.

Is This Answer Correct ?

9 Yes

5 No

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / kailas.s.patil

i = 2;
first ++i = 3.
second ++i = 4.
third i++ = 5;
now i =5;

then, j = 3 + 4 + 5 = 12

Is This Answer Correct ?

4 Yes

1 No

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / kapil

i=5
j=11

j=3+4+4; (i++ post increment)
after this step i value become 5(i++)

Is This Answer Correct ?

2 Yes

0 No

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / sanjay

i = 5
j = 11
It is because during the first pre-increment "++i" the compiler gets the value from the memory, increments it and stores it in the memory ie now i = 3. During the second pre-increment "++i" the compiler again gets the value from the memory, increments it, (value in the memory was 3) and so the incremented value is stored again in memory ie i = 4. during the post increment, the value from the memory is received and used in the statement ie) (the whole final statement looks like this ->>( 3 + 4 + 4) ) and then value of i is incremented and stored in memory. thus finally the value of i is 5 and j is 11.

Is This Answer Correct ?

2 Yes

0 No

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / vadim g

i=5 and j=12

Here is the sequence of operations:
1. i = 2 => result i = 2
2. first ++i => result: i = 3
3. second ++i => result: i = 4
4. j assignment => result: j = 4 + 4 + 4 = 12
5. i post increment => i = 5

MSVC++ gives 5 and 12.

Is This Answer Correct ?

4 Yes

3 No

#include<stdio.h> int main() { int i=2; int j=++i + ++i + i++; printf("%d\n&qu..

Answer / lebin

value of i is 5
value of j is 12

Is This Answer Correct ?

3 Yes

2 No

Post New Answer