From: Farrel Lifson <farrel.lifson@...>
Date: 2006-08-07T23:53:47+09:00
Subject: Re: nextPowerOf2(n)

On 07/08/06, Farrel Lifson <farrel.lifson@gmail.com> wrote:
> On 07/08/06, Farrel Lifson <farrel.lifson@gmail.com> wrote:
> > On 07/08/06, Ch Skilbeck <ruby@skilbeck.com> wrote:
> > > Hi,
> > >
> > > Can someone tell me if there's a better way to do this? It takes a
> > > number and returns the next power of 2 up (or the original if it was
> > > already a power of 2) Specifically, are there features of Ruby that I
> > > should be using in this case?
> > >
> > > Cheers,
> > > Charlie.
> > >
> > > def nextPowerOf2(n)
> > >   raise "integer please" if !n.kind_of? Integer
> > >   high = 0
> > >   count = 0
> > >   (0..n.size * 8 - 2).each {|b| count += n[b] ; high = b if n[b] != 0}
> > >   1 << high + (count > 1 ? 1 : 0)
> > > end
> > >
> > >
> > > --
> > > Posted via http://www.ruby-forum.com/.
> > >
> > >
> >
> > My attempt:
> >
> > # Finds the log base 2 of a number
> > def Math.log2(n)
> >   Math.log(n)/Math.log(2)
> > end
> >
> > def nextPowerOf2(n)
> >   if Math.sqrt(n).modulo(1).zero?
> >     n
> >   else
> >     2**Math.log2(5).to_i.succ
> >   end
> > end
> >
> >  Farrel
> >
>
> Gah! I don't even need that Math.log2 function
>
> def nextPowerOf2(n)
>   if Math.sqrt(n).modulo(1).zero?
>     n
>   else
>     2**Math.sqrt(5).to_i.succ
>   end
> end
>
> Farrel
>
Sorry for the repeated typos, it shoud be '2**Math.sqrt(n).to_i.succ'