[Haskell-cafe] "show" for functional types

[Neil Mitchell]

> > now superShow f /= superShow g, so they are no longer referentially transparent.
>
>     OK.  I'm probably being really dense today, but where did "g" come
> from?
Nope, I was being super dense, for "g" read "not"...

superShow f = "not"
superShow not = "\x -> case x of {True -> False; False -> True}"

superShow f /= superShow not


> And does this loss
> of referential transparency contaminate the rest of the language, or is
> this superShow just an anomaly?
Once you have any loss of transparency, you loose it totally. You
would then have to go about proving that superShow was not called
anywhere in the vicinity of the transformation you wish to appy,
inlining by the compiler would no longer be safe...

Thanks

Neil