[Haskell-cafe] New to Haskell
Cristian Baboi
cristi at ot.onrc.ro
Tue Dec 18 02:31:59 EST 2007
A few days ago, for various reasons, I've started to look at Haskell.
At first I was quite impressed, after reading some FAQ, and some tutorials.
Evrything was nice and easy ... until I've started writing some code on my
own.
What I should have been told about upfront:
- the syntax for an expression
- the syntax for a block
- the adhoc syntax rules (how to distinguish among a tuple and a
pharanthesized expression and how to find the start and end of a block for
example )
- the fact that lambda expressions are not the same thing as "algebraic
data" values
- what guarantees are made by the LANGUAGE that an IO action (such as do
putStrLn "Hello world" ) is not performed twice
- the lambda expressions can be written (input) but cannot be printed
(output)
The biggest problem for me, so far, is the last one.
Here is some strange example:
module Hugs where
aa::Int
aa=7
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 }
f::Int->Int
f(1)=1
f(2)=2
f(_)=3
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
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) and why in function h, the bounded u and
v become free variables in the case expression.
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