[Haskell-cafe] New to Haskell

Ketil Malde ketil+haskell at ii.uib.no
Tue Dec 18 03:39:46 EST 2007

"Cristian Baboi" <cristi at ot.onrc.ro> writes:

> Here is some strange example:

> module Hugs where

> aa::Int
> aa=7

Small note, it's common to use spaces around the :: and =   I've
never really noticed before.

> cc :: (Int->Int) -> (Int->Int->Int) -> Int -> (Int->Int)
> cc a op b  = \x-> case x of { _ | x==aa -> x+1 ;  _-> a x `op` b }

> What I don't understand is why I'm forced to use guards like x==aa in
> cc,  when aa is clearly bounded (is 7) 

I don't quite understand what you mean.  You don't have to use guards,
the function could equally well have been written using if-then-else.
Why not

  cc a op b x = if x==aa then (x+1) else a x `op` b

Oh, wait, you're asking why you can't write

    case x of aa -> x+1
              _  -> a x `op` b

The answer is that case introduces a new binding for 'aa', so the
above is equivalent to 

      let aa = x in x+1

Case is really for deconstructing values with pattern matching, a
simple variable like aa (or _) will match any pattern.

> f::Int->Int
> f(1)=1
> f(2)=2
> f(_)=3

You can drop the parentheses here.

> g::Int->Int
> g(1)=13
> g(2)=23
> g(_)=33

> h :: [Int->Int] -> Int -> Int
> h []   x = x
> h [rr] x =  let { u=Hugs.f ; v=Hugs.g } in  case rr of  {  u  ->
> Hugs.g(x)+aa ; v -> Hugs.f(x)+aa ; _ ->rr (x) + aa }
> h  (rr:ll)  x =  h [rr] x + h (ll) x

Same here, if I understand you correctly.  The case introduces new
bindings for u and v.  Note that you can't (directly) compare
functions for equality either, the only way to do that properly would
be to compare results over the entire domain.  (f == g iff f x == g x
forall x)

If I haven't seen further, it is by standing in the footprints of giants

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