Re: from number to power of two

[Kazufumi-MIT-Mitani]

 "Daniel C. Sobral" <[EMAIL PROTECTED]> wrote

> That technique is O(ln(n)), where n is the number in question.
> 
> Frankly, for numbers up to 32, a table will wield the best results,
> and might actually be smaller than some of the suggestions given so
> far.

Counting n as bit, it is O(n) :p

Unrolling the loop,

if(n<=2^0) 2^0
else if(n<=2^1) 2^1
else if(n<=2^2) 2^2
else if(n<=2^3) 2^3
   :
   :
else if(n<=2^31) 2^31

is suitable for this problem?

---
//mit



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