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main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}

Answers were Sorted based on User's Feedback



main() { int i=0; while(+(+i--)!=0) i-=i++; printf("%d",i); }..

Answer / furquan

Here what happens is :

while(+(+i--)!=0)
i = 0
so 0 != 0 is false hence condition for while fails. Now
there is also a post decrement operator. So value of i turns
out to be -1. and that is which is printed.

Is This Answer Correct ?    6 Yes 0 No

main() { int i=0; while(+(+i--)!=0) i-=i++; printf("%d",i); }..

Answer / rameshp

ans s -1

Bec

while(1!=0) this s false..so next line don't exec..

so i-- means 0--= -1

finally we get -1

Is This Answer Correct ?    6 Yes 1 No

main() { int i=0; while(+(+i--)!=0) i-=i++; printf("%d",i); }..

Answer / priyadarshan kasta

-1.


i don't know why.

plz post the reason

Is This Answer Correct ?    3 Yes 1 No

main() { int i=0; while(+(+i--)!=0) i-=i++; printf("%d",i); }..

Answer / rayan

ans is -1.

Initially i == 0, when it enters into while loop

while(+(+i--)!0)

in 1st iteration i value ll be 0 only due to post decrement
operator.

in second iteration as i gets its post decremented value n
becomes -1 & -1 ~= 0 condition gets true & while loop breaks.

here + operator in the loop is doing ntg.

Is This Answer Correct ?    2 Yes 1 No

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