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From: Willem <willem@toad.stack.nl>
Newsgroups: comp.theory,comp.programming,sci.math
Subject: Re: n bits on of m bits
Date: Thu, 22 Dec 2011 15:58:19 +0000 (UTC)
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Robin Vowels wrote:
) On Dec 18, 3:05?am, Andrew Tomazos <and...@tomazos.com> wrote:
)> I want to loop over every number from 0 to (2^m-1) inclusive whos
)> binary representation contains exactly n bits.
)>
)> ie
)>
)> for (int i = 0; i < (1<<m); i++)
)> {
)> ? ? int bits = 0;
)> ? ? for (int j = 0; j < m; j++)
)> ? ? ? ? if (i & (1 << j))
)> ? ? ? ? ? ? bits++;
)> ? ? if (bits == n)
)> ? ? ? ? do_something_with(i);
)>
)> }
)>
)> Clearly this is terribly inefficient. ?I'm sure there is a much faster
)> way (perhaps by enumerating with mod/div somehow each continous
)> sequence of 1s and continuous sequence of 0'x) but I can't seem to see
)> it.
)
) It is unclear what you want.

Read the code then.  It does what the OP wants, it's just very slow.

) Or do you mean n one-bits (as someone has assumed)?

Not assumed, inferred.  And not someone, everyone (except you).

HTH, HAND.


SaSW, Willem
-- 
Disclaimer: I am in no way responsible for any of the statements
            made in the above text. For all I know I might be
            drugged or something..
            No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT