Hi Friends,

Here is another interesting C Problem.
The problem is to find whether an integer is a power of 2 or not.
This must be done in a single C statement.
I tried a little but have not figured it out as yet.


I have been thinking on the following lines:

I suppose it has something to do with bit manipulation coz.. all integers that are power of 2 will have only single 1 in binary form, we need to be able to count if it has only single one..

Is there ny way???

Thanx.

So easy.
#include <cmath>
take a log of base 2 //don't remember the function, look it up on google
and take an integer test.

That should do it

Thanks. that is a good idea...
i'll check it out...
btw keep posted if u get an idea on how to do it without using built-in functions.

that's not hard either

Do a repeat x <= 1
{
number / 2;
x -= 1;
}

if x==1, then it's a power of two.

If you want optimization, well... specify.

hi osknockout,

That is a simple solution any one wud think of as first option...
but the condition in the question is not satisfied here. i.e., "it shoud be determined in a single C Statement"

so what i mean is some other solution in a single C Statement which does not use built in function...
the solution u have give has more than that....


Regards.

Right, sorry, I'll post if I ever find out...

So the solution set is:
00000010
00000100
00001000
00010000
00100000
01000000
10000000
ad infinitum

I think this is more easy:

#define IS_2_POWER(X) (X==(~X)+1)

Nice idea but...

00000010 = 2
11111101 = ~2
11111110 = ~2 + 1

x==(~X)+1?
don't think so.

Good attempt though

I've GOT IT! (Thanks yomi)


Ok, I'll work on a more optimized formula for right now...

oh, i made a mistake , think about this:
x==((~x)+1)|x

I am not very sure. But sure there is a simular way to achieve this operation with high performance.