Re: [Haskell-cafe] Newbie question on Haskell type

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@bloglines.com wrote:

--- Cale Gibbard

...

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.