main() {int i=5; // line 1 i=(++i)/(i++); // line 2 printf("%d",i);

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?

Question Posted / sandeep gupta

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main() {int i=5; // line 1 i=(++i)/(i++); // line 2 printf("%d",i); // line 3 } output..

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.

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