[Haskell] instance Bounded Double
Lennart Augustsson
lennart at augustsson.net
Sun Mar 13 18:02:35 EST 2005
And what would you have minBound and maxBound be?
I guess you could use +/- the maximum value representable.
Going for infinity is rather dodgy, and assumes an FP
representation that has infinity.
-- Lennart
Frederik Eaton wrote:
> Interesting. In that case, I would agree that portability seems like
> another reason to define a Bounded instance for Double. That way users
> could call 'maxBound' and 'minBound' rather than 1/0 and -(1/0)...
>
> Frederik
>
> On Fri, Mar 11, 2005 at 11:10:33AM +0100, Lennart Augustsson wrote:
>
>>Haskell does not guarantee that 1/0 is well defined,
>>nor that -(1/0) is different from 1/0.
>>While the former is true for IEEE floating point numbers,
>>the latter is only true when using affine infinities.
>>
>> -- Lennart
>>
>>Frederik Eaton wrote:
>>
>>>Shouldn't Double, Float, etc. be instances of Bounded?
>>>
>>>I've declared e.g.
>>>
>>>instance Bounded Double where
>>> minBound = -(1/0)
>>> maxBound = 1/0
>>>
>>>in a module where I needed it and there doesn't seem to be any issue
>>>with the definition...
>>>
>>>Frederik
>>>
>>
>>
>
More information about the Haskell
mailing list