Hal Daume III wrote:
> This works great for when x/=0...is there a good (Haskell) solution for
> the smallest positive float?
I think that the following are correct for the smallest normalised
Float and Double values:
Prelude> encodeFloat 1 (fst (floatRange (0 :: Float)) - 1) :: Float
1.1754944e-38
Prelude> encodeFloat 1 (fst (floatRange (0 :: Double)) - 1) :: Double
2.2250738585072014e-308
They (roughly) agree with FLT_MIN and DBL_MIN from <float.h>:
#define FLT_MIN 1.17549435e-38F
...
#define DBL_MIN 2.2250738585072014e-308
Furthermore, these values aren't denormalised, but reducing the
exponent by one gives a denormalised value:
Prelude> isDenormalized (encodeFloat 1 (fst (floatRange (0 :: Float)) - 1) ::
Float)
False
Prelude> isDenormalized (encodeFloat 1 (fst (floatRange (0 :: Float)) - 2) ::
Float)
True
Prelude> isDenormalized (encodeFloat 1 (fst (floatRange (0 :: Double)) - 1) ::
Double)
False
Prelude> isDenormalized (encodeFloat 1 (fst (floatRange (0 :: Double)) - 2) ::
Double)
True
These appear to give the smallest possible Float/Double values:
Prelude> encodeFloat 1 (fst (floatRange (0 :: Double)) - floatDigits (0 ::
Double)) :: Double
5.0e-324
Prelude> encodeFloat 1 (fst (floatRange (0 :: Float)) - floatDigits (0 ::
Float)) :: Float
1.0e-45
Reducing the exponent by 1 gives 0.0:
Prelude> encodeFloat 1 (fst (floatRange (0 :: Float)) - floatDigits (0 ::
Float) - 1) :: Float
0.0
Prelude> encodeFloat 1 (fst (floatRange (0 :: Double)) - floatDigits (0 ::
Double) - 1) :: Double
0.0
--
Glynn Clements <[EMAIL PROTECTED]>
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