Re: [Haskell] instance Bounded Double

2005-03-14 Thread Stefan Karrmann
John Meacham (Sun, Mar 13, 2005 at 08:08:56PM -0800): On Sun, Mar 13, 2005 at 11:08:26PM +, Thomas Davie wrote: [...] We could define maxBound as (2^(mantisa_space))^(2^(exponent_space)) and min bound pretty similarly... But I'm sure that everyone will agree that this is a horrible

Re: [Haskell] instance Bounded Double

2005-03-13 Thread Frederik Eaton
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

Re: [Haskell] instance Bounded Double

2005-03-13 Thread Lennart Augustsson
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

Re: [Haskell] instance Bounded Double

2005-03-13 Thread Thomas Davie
I may be barking up the wrong tree here, but I think the key to this discussion is that real numbers are not bounded, while doubles are bounded. One cannot say what the smallest or largest real number are, but one can say what the smallest or largest double are (and it is unfortunately

Re: [Haskell] instance Bounded Double

2005-03-13 Thread Lennart Augustsson
I agree with all of that. :) -- Lennart Thomas Davie wrote: I may be barking up the wrong tree here, but I think the key to this discussion is that real numbers are not bounded, while doubles are bounded. One cannot say what the smallest or largest real number are, but one can say what

Re: [Haskell] instance Bounded Double

2005-03-13 Thread John Meacham
On Sun, Mar 13, 2005 at 11:08:26PM +, Thomas Davie wrote: I may be barking up the wrong tree here, but I think the key to this discussion is that real numbers are not bounded, while doubles are bounded. One cannot say what the smallest or largest real number are, but one can say what

Re: [Haskell] instance Bounded Double

2005-03-11 Thread Lennart Augustsson
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

[Haskell] instance Bounded Double

2005-03-10 Thread Frederik Eaton
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 -- http://ofb.net/~frederik/