following using exor
http://www.geeksforgeeks.org/archives/18324
following is tricky
http://www.geeksforgeeks.org/archives/22080
On Wed, Nov 7, 2012 at 11:36 AM, Rahul Kumar Patle <
patlerahulku...@gmail.com> wrote:
> you can use bit wise addition.. using xor , and , or and shift op
> erations..
let me keep it simple
y is integer pointer pointing to address 20 .. x=20
y+7 will point to the addressy + ( size(int) * 7) i.e 50 here
/*y is integer pointer*/
On Tue, Nov 6, 2012 at 11:45 AM, Rahul Kumar Patle <
patlerahulku...@gmail.com> wrote:
> because difference is of 30 byt
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 --
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;
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 becau
