off topic but please forgive me me and answer

[Tim Chase]

Steven D'Aprano wrote:
>> That's because you need to promote one of them to a float so you get a
>> floating-point result:
>>
>>    >>> 1/2 * 1/2
>>    0
>>    >>> 1/2 * 1/2.0
>>    0.0
>>
>> Oh...wait ;-)
> 
> Tim, I'm sure you know the answer to this, but for the benefit of the 
> Original Poster, the problem is that you need to promote *both* divisions 
> to floating point. Otherwise one of them will give int 0, which gives 0.0 
> when multiplied by 0.5.
> 
>>>> 1.0/2 * 1/2.0
> 0.25

You can get away with just promoting one of them...you just have 
to promote the _correct_ one (one involved in the first division) 
so that its promotion-of-subresult-to-float carries into all 
subsequent operations/operators:

   >>> 1/2 * 1/2 # (((1/2)*1)/2)==(((0)*1)/2) in 2.x
   0
   >>> 1/2 * 1/2.0 # (((1/2)*1)/2.0)==(((0)*1)/2.0) in 2.x
   0.0
   >>> 1/2 * 1.0/2 # (((1/2)*1.0)/2)==(((0)*1.0)/2) in 2.x
   0.0
   >>> 1/2.0 * 1/2 # (((1/2.0)*1)/2)
   0.25
   >>> 1.0/2 * 1/2 # (((1.0/2)*1)/2)
   0.25

I'd rather be explicit in *real* code that I'd write and 
explicitly float'ify constants or float() integer variables. The 
OP's question was both OT and pretty basic middle-school math 
that google would have nicely answered[1] so IMHO warranted a bit 
of fun. :)

-tkc

[1]
http://www.google.com/search?q=1%2F2+*+1%2F2