[Haskell-cafe] Re: instance Eq (a -> b)

[Ashley Yakeley]

On 2010-04-15 06:18, Nick Bowler wrote:

> Your definitions seem very strange, because according to this, the
> functions
>
>    f :: Double ->  Double
>    f x = 1/x
>
> and
>
>    g :: Double ->  Double
>    g x = 1/x
>
> are not equal, since (-0.0 == 0.0) yet f (-0.0) /= g (0.0).

There's an impedance mismatch between the IEEE notion of equality (under 
which -0.0 == 0.0), and the Haskell notion of equality (where we'd want 
x == y to imply f x == f y).

A Haskellish solution would be to implement Eq so that it compares the 
bits of the representations of Float and Double, thus -0.0 /= 0.0, NaN 
== NaN (if it's the same NaN). But this might surprise people expecting 
IEEE equality, which is probably almost everyone using Float or Double.

-- 
Ashley Yakeley