Using while loop and if statement to tell if a binary has an odd or even number of 1's.

[MRAB]

Mark Dickinson wrote:
> On Feb 5, 1:18 am, Chris Rebert <c... at rebertia.com> wrote:
>> For an integer:
>> is_even = bin(the_int)[2:].count('1') % 2 == 0
> 
> But the OP has to use if and while.  How about:
> 
> while 2+2 != 5:
>     if 'wkw' in 'just being awkward':
>         is_even = bin(the_int)[2:].count('1') % 2 == 0
>         break
> 
> or (Python 2.5 compatible):
> 
> def count_set_bits(n):
>     # make sure we include an if, to
>     # satisfy OP's requirements:
>     if n < 0:
>         raise ValueError
>     count = 0
>     while n:
>         count += 1
>         n &= n-1
>     return count
> 
> is_even = count_set_bits(the_int) % 2 == 0
> 
> ...but anyone submitting this as a homework
> solution had better be prepared to explain why
> it works.
> 
I haven't seen that algorithm before. Very clever! :-)