Thanks! I figured I was close. Didn't even know const was available. I put together a compliment functions earlier complement :: (a -> Bool) -> a -> Bool complement p x = not (p x) By the signature, the first argument is a function (predicate) which when given a value returns a Bool? And the second argument is just a value? And the function returns a Bool?
map (complement odd) [1,2,3,4,5,6] [False,True,False,True,False,True]
By similar reasoning the always function would seem to
have a signature
a -> (b -> a)
where the first argument is just a value and the
return value is a function that when given a possibly
different value just returns the value originally
given to always?
Is that reasoning OK? Are
a -> (b -> a) and a -> b -> a the same signature?
So the inferred type is usually pretty accurate? These
signatures are a bit confusing. Is there a good
tutorial?
I'm using Hugs/Win XP just to scope out the language
right now. I tried what you suggested and got
Hugs> let always x _ = x
ERROR - Syntax error in expression (unexpected end of
input)
Hugs>
Isn't Hugs an interpreter?
Thanks again. Really interesting language Haskell.
Michael
--- Matthew Brecknell
This is what I've been trying:
always :: (a -> a) -> a -> a always x = (\y -> x)
Your function implementation is correct, but the type is wrong. Try this:
always :: a -> b -> a
Or, just use the function "const", from the Prelude. :-)
The type system can be very handy when learning Haskell. If you think you have the correct implementation but can't work out the type, just start up an interpreter and ask it for the inferred type. For example:
Prelude> let always x _ = x Prelude> :t always always :: t -> t1 -> t
Once you have the type, ask Hoogle if the function already exists:
http://haskell.org/hoogle/?q=t+-%3E+t1+-%3E+t
And there is "const" at the top of the results. :-)
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