I'm trying to learn Haskell and translating some Lisp
functions as exercises.
How would I write a Haskell function named ALWAYS that
behaves like this:
one = always 1
bozo = always "clown"
> map one [2,3,4,5,6]
[1,1,1,1,1]
> one 62
1
> map bozo [2,3,4,5,6]
["clown","clown" ,"clown", "clown"," clown"]
> bozo 62
"clown"
i.e. ALWAYS returns a function with a single parameter
that is ignored, returning instead the value given to
ALWAYS when the function was created.
This is what I've been trying:
always :: (a -> a) -> a -> a
always x = (\y -> x)
one = always 1
Michael
First, you want 'one' to take an integer and return 1. So,
one :: Integer -> Integer
since one = always 1, then
always 1 :: Integer -> Integer
So, 'always' takes an Integer and returns an Integer -> Integer
always :: Integer -> (Integer -> Integer)
But that's the same as
always :: Integer -> Integer -> Integer
You actually have the implementation correct, you just didn't have the
right type signature.
always first = (\second -> first)
Of course, neither of these implementations need to be tied to
Integers; they can be polymorphic. So, we end up with:
always :: a -> b -> a -- no reason the second parameter has to be the
same type as the first, so use 'b' instead of 'a'.
always first = (\_ -> first) -- replace 'second' with '_', because
we don't need to bind anything to the second parameter.
Does that makes sense?
Bryan Burgers
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