We start with k = 5.
if ++k < 5 translates to if 6 < 5 => which is false.
No need to evaluate k++/5 (short-circuiting)
if ++k <= 8 translates to (6+1) 7 <= 8 which is true.
So, 7 gets printed.
On Tuesday, November 6, 2012 6:22:13 AM UTC-5, Anil Sharma wrote:
>
> main()
> {
> int k = 5;
> if (++k < 5 && k++/5 || ++k <= 8);
> printf("%d ", k);
> }
>
> the output shud be 8 but it comes out to be 7.why???
> as increment operator has higher precedence among them so increment shud
> be done throughout at first and after then other operators shud be
> evaluated.so output shud be 8.
>
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To view this discussion on the web visit
https://groups.google.com/d/msg/algogeeks/-/KgsSjel6n_4J.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
