Log base 2 of large integers

[Peter Otten]

Steven D'Aprano wrote:

> Mok-Kong Shen wrote:
> 
>> 
>> I like to compute log base 2 of a fairly large integer n but
>> with math.log(n,2) I got:
>> 
>> OverflowError: long int too large to convert to float.
>> 
>> Is there any feasible work-around for that?
> 
> If you want the integer log2, that is, the floor of log2, the simplest way
> is calculate it like this:
> 
> def log2(n):
>     """Return the floor of log2(n)."""
>     if n <= 0: raise ValueError
>     i = -1
>     while n:
>         n //= 2
>         i += 1
>     return i
> 
> log2(511)
> => returns 8
> log2(512)
> => returns 9
> log2(513)
> => returns 9

For base 2 there is also the bit_length() method:

>>> 511 .bit_length()
9
>>> 512 .bit_length() 
10
>>> 513 .bit_length()  
10