Let's concentrate on if condition,
*if (++k < 5 && k++/5 || ++k <= 8);*
is equivalent to (based on operator precedence)
*if (((++k < 5) && (k++/5)) || (++k <= 8));*
Initially k = 5.
Step1 : The first term in if clause i.e, (++k < 5) is false which makes the
*((++k < 5) && (k++/5)) *false because all the variables in && should be 1.
this makes k = 6 because of execution of (++k < 5).
Step 2 : Now if condition is equivalent to if(0 || (++k <= 8))
it will check the second condition, (++k <= 8) thereby incrementing k by 1
and returns 1 (making if condition true though it's not relevant for this
problem)
Hence, k = 7 after execution of this program.
On Tuesday, November 6, 2012, 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.
>
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Aamir Khan | 4th Year | Computer Science & Engineering | IIT Roorkee
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