test whether a number is a power of 2

[Matt]

Give a one-line C expression to test whether a number is a power of 2.
No loops allowed.

[Joona I Palaste]

Matt <jr********@hot mail.com> scribbled the following:

Give a one-line C expression to test whether a number is a power of 2.
No loops allowed.

((((x&&!(x&(x-!!x)))+x)-x)^x)^x

What do I win?

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/-- Joona Palaste (pa*****@cc.hel sinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
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"Remember: There are only three kinds of people - those who can count and those
who can't."
- Vampyra

[Ivan Vecerina]

"Matt" <jr********@hot mail.com> wrote in message
news:ba******** *************** ***@posting.goo gle.com...

Give a one-line C expression to test whether a number is a power of 2.
No loops allowed.

isPow2 = x && !( (x-1) & x );

See http://www.caam.rice.edu/~dougm/twiddle/
for some related info.

hth
--
http://ivan.vecerina.com
http://www.brainbench.com <> Brainbench MVP for C++

[Keith Thompson]

jr********@hotm ail.com (Matt) writes:

Give a one-line C expression to test whether a number is a power of 2.
No loops allowed.

We're not here to do your homework for you.

--
Keith Thompson (The_Other_Keit h) ks*@cts.com <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://www.sdsc.edu/~kst>
Schroedinger does Shakespeare: "To be *and* not to be"

[jacob navia]

isPow2 = x && !( (x-1) & x );

Algorithm:
If x is a power of two, it doesn't have any bits in common with x-1, since it
consists of a single bit on. Any positive power of two is a single bit, using
binary integer representation.

This means that we test if x-1 and x doesn't share bits with the and operator.

Of course, if x is zero (not a power of two) this doesn't hold, so we add an
explicit test for zero with xx && expression.

Negative powers of two (0.5, 0.25, 0.125, etc) could share this same property
in a suitable fraction representation. 0.5 would be 0.1, 0.250 would be 0.01,
0.125 would be 0.001 etc.

jacob

[Aishwarya]

Joona I Palaste <pa*****@cc.hel sinki.fi> wrote in message news:<bl******* ***@oravannahka .helsinki.fi>.. .

Matt <jr********@hot mail.com> scribbled the following:

Give a one-line C expression to test whether a number is a power of 2.
No loops allowed.

((((x&&!(x&(x-!!x)))+x)-x)^x)^x

What do I win?

hi Joona,

i know this works because i just tested it .. but help me understand
the algoritm please ..

Thanks and Regards
A.

[Eric Sosman]

Matt wrote:

Give a one-line C expression to test whether a number is a power of 2.
No loops allowed.

true_or_false = a_number > 0;

(For example, 3.14159 ~= 2 ** 1.65149.)

--
Er*********@sun .com

[Dan Pop]

In <64************ *************@p osting.google.c om> ai*******@sapie nt.com (Aishwarya) writes:

Joona I Palaste <pa*****@cc.hel sinki.fi> wrote in message news:<bl******* ***@oravannahka .helsinki.fi>.. .

Matt <jr********@hot mail.com> scribbled the following:

> Give a one-line C expression to test whether a number is a power of 2.
> No loops allowed.

((((x&&!(x&(x-!!x)))+x)-x)^x)^x

i know this works because i just tested it .. but help me understand
the algoritm please ..

Had you engaged your brain, you'd have easily noticed that one ^x cancels
the other and that -x cancels the +x, so you're left with

x&&!(x&(x-!!x))

Now, !!x is an obfuscated way of writing 1, when you know that x cannot
be zero (it was already tested), so, we're left with:

x&&!(x&(x-1))

The tricky bit is x&(x-1) which actually clears the least significant
bit that happens to be set. If the original number was a power of two,
it had only one bit set, so the result of x&(x-1) must be zero in such
cases.

The rest should be obvious.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Dan Pop]

In <lz************ @cts.com> Keith Thompson <ks*@cts.com> writes:

jr********@hot mail.com (Matt) writes:

Give a one-line C expression to test whether a number is a power of 2.
No loops allowed.

We're not here to do your homework for you.

OTOH, the solution is far too tricky for a homework assignment. It
requires a bit of knowledge that has precious little to do with C
programming.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Jirka Klaue]

Aishwarya wrote:

Joona I Palaste <pa*****@cc.hel sinki.fi> wrote in message news:<bl******* ***@oravannahka .helsinki.fi>.. .

....

((((x&&!(x& (x-!!x)))+x)-x)^x)^x

i know this works because i just tested it .. but help me understand
the algoritm please ..

+x -x does nothing, ^x ^x does nothing.

remains:
x && !(x & (x - !!x))

!!x ist alway 1, because x can't be 0 at this point.

remains:
x && !(x & (x - 1))

Jirka