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