[Haskell-cafe] Newbie question on Haskell type

[Cale Gibbard]

Right, forgot about seq there, but the point still holds that there
are a very limited number of functions of that type, and in
particular, the functions can't decide what to do based on the type
parameter 'a'.

 - Cale

On 14 Oct 2005 05:49:27 -0000, voigt.16734551 at bloglines.com
<voigt.16734551 at bloglines.com> wrote:
> --- Cale Gibbard <cgibbard at gmail.com wrote:
>
> > As an example of this sort
> of thing, I know that there are only 4
> > values of type a -> Bool (without
> the class context). They are the
> > constant functions (\x -> True), (\x ->
> False), and two kinds of
> > failure (\x -> _|_), and _|_, where _|_ is pronounced
> "bottom" and
> > represents something along the lines of nontermination (aborting
> the
> > program also counts).
>
> Not exactly. There are also (\x -> seq x True)
> and (\x -> seq x False), neither of which is equivalent to any of the four
> functions above.
>
> Regards,
>
> Janis Voigtlaender.