lambda with floats

[monkeys paw]

On 4/7/2010 1:08 PM, Peter Pearson wrote:
> On Tue, 06 Apr 2010 23:16:18 -0400, monkeys paw<monkey at joemoney.net>  wrote:
>> I have the following acre meter which works for integers,
>> how do i convert this to float? I tried
>>
>> return float ((208.0 * 208.0) * n)
>>
>>>>> def s(n):
>> ... 	return lambda x: (208 * 208) * n
>> ...
>>>>> f = s(1)
>>>>> f(1)
>> 43264
>>>>> 208 * 208
>> 43264
>>>>> f(.25)
>> 43264
>
> The expression "lambda x: (208 * 208) * n" is independent of x.
> Is that what you intended?
>
>

Seems i should have done this:
g = lambda x: 208.0 * 208.0 * x
g(1)
43264.0