void main() { int i=5; printf("%d",i++ + ++

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void main()
{
int i=5;
printf("%d",i++ + ++i);
}

Question Posted / surenda pal singh chouhan

21 Answers
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void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / ishrat ali

the calculate of any expression can be done from left to
right,so first i++ compute it means 1st use then increment 5
is used in place of i++ and after using the value of i
increment that is 6.
after that ++i is used that is 1st increment then use in the
expression so now ++i is replaced by 7;
finally 6+7=12 (ans)

Is This Answer Correct ?

1 Yes

0 No

void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / veeraselvi.g

ans:12
explan:i++ means increment for 6 bt print will be 5;
other i++ for 6 to increment for 7
so ans is 12

Is This Answer Correct ?

0 Yes

0 No

void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / gajendra

Is This Answer Correct ?

1 Yes

1 No

void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / poseidon

I might agree with ANS #1 because side effects are
involved..The expected answer might not be getting always
although in case of printf(), the arguments are evaluated
from RTL. For example consider the case

int i=10;
printf("%d %d %d %d",i++,--i,i--,--i);
Try some variations of these and and find out whether you
are able to get correctly or not..
In case the result depends on the compiler,for
verification,I will add what my compiler gave me
ANS:7 8 9 8
If anyone find out,pls reply.. Keen to know..

Is This Answer Correct ?

0 Yes

0 No

void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / buvaneswari

as already told,the precedence wil be from right to left..
so i++ the value wil be still 5 and incremented value wil be stored in memory..
and next ++i now the value of i=5 not considering the memory value...so ++i =6...now the new value of i is 6.this wil be replaced the value 5 in the memory
u can see the expression as i+i in default or consider to be
i=i++ and i=++i....so i+i(ie)6+6=12....

Is This Answer Correct ?

0 Yes

1 No

void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / pnkj rajput@kiit,bbsr

Ans. is

12

because

(first check) i++ (no change in the value of 'i' so i=5)

(second check) ++i ('i' increase by 1 so, i=6)

Now, in this the value of 'i' is 6

so, 6 + 6 = 12

Is This Answer Correct ?

0 Yes

1 No

void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / kaushik

13

i=i++--->6

++6=7

7+6=13

Is This Answer Correct ?

0 Yes

1 No

void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / surenda pal singh chouhan

Output Cannot be predicted exactly.

Explanation:
Side effects are involved in the evaluation of i

Is This Answer Correct ?

13 Yes

15 No

void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / guest

same as the pervious question....

11 ans

Is This Answer Correct ?

11 Yes

22 No

void main() { int i=5; printf("%d",i++ + ++i); }..

Answer / jignesh patel

i++=5
++i=6

so i++ + ++i = 5+6=11

Is This Answer Correct ?

6 Yes

26 No

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