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main()
{int i=5; // line 1
i=(++i)/(i++); // line 2
printf("%d",i); // line 3
} output is 2 but if we replace line 2 and line 3 by
printf("%d",i=(++i)/(i++)); then output is 1. Why?
- 1 Answers
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Answer / gagandeep bansal
working of incr/decr operator:first pre operators then
rest of operators then post operators
so,2 line become:i=6/(5++); /*pre operator*/
i=6/5=1; /*rest of operators*/
now post incr operator will work so,i=2; /*post operator*/
so output is 2.
in second case:i=6/(5++); /*pre operator*/
i=6/5==1; /*rest of operators*/
now post incr operator will work so,i=2;
again i=3/(2++);
i=1;
so,output is 1.
becouse in case of post operators first assign then
incr/decr.
| Is This Answer Correct ? | 9 Yes | 3 No |
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