I guess its other way.
According you guys explanation after evaluating the expression first the if
condition turns out to be if(1);
And the program runs infinitely..
first -- ++k <= 8 --- turns true and k becomes 6.
second -- k++ / 5 --- skipped. k remains 6
third -- ++k < 5 --- turns false and k becomes 7.
Hope its very clear
On Wed, Nov 7, 2012 at 7:13 AM, apsalar <kavya...@gmail.com> wrote:
> 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.
>>
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