haskell error function

Question

I have one function first with type: Int -> [a] -> (Error ([a],[a])) and a second function second with type: [a] -> [a] -> [a]

I am trying to make a third function now that uses the above functions. the type I have for this function is: [Int] -> [a] -> Error [a]

I have been given these types to work around so cant change them.

This is what I tried:

last :: [Int] -> [a] -> Error [a]
last (x:xs) list = second (first x list)

Can you pass outputs from functions that use the error function in to others?

Function names must be lower case. Have you tried implementing this? — Joel Burget
– Joel Burget, Commented Mar 24, 2011 at 15:43
@Joel you can have functions with upper case names. They're also known as data constructors (ok, maybe function-like values?). E.g. Just :: a -> Maybe a. — R. Martinho Fernandes
– R. Martinho Fernandes, Commented Mar 24, 2011 at 15:44
I understand that data constructors are upper case, but F and Q are not data constructors. — Joel Burget
– Joel Burget, Commented Mar 24, 2011 at 15:47
i have changed the function names, that was just for the example, fixed. — Lunar
– Lunar, Commented Mar 24, 2011 at 17:52
What is Error here? The answer from Martinho seems to imply that it is a monad, is it so? — Ed'ka
– Ed'ka, Commented Mar 24, 2011 at 22:50

R. Martinho Fernandes · Accepted Answer · 2011-03-25 09:41:29Z

3

Assuming Error is an error monad, you can use the monadic bind operator (>>=) and the uncurry function:

z (x:xs) list = F x list >>= return . uncurry Q

uncurry transforms Q from a function with two arguments (aka a curried function) into a function on pairs. This means that uncurry Q :: ([a], [a]) -> [a]

The bind operator takes a value out of a monad and passes it into a monadic function. Here we're extracting the value of the Error monad returned by F and passing it to Q, now turned into a monadic function that works on a pair of lists, thanks to return and uncurry.

edited Mar 25, 2011 at 9:41

answered Mar 24, 2011 at 15:51

R. Martinho Fernandes

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3 Comments

fuz Over a year ago

You can also use the functor instance instead: z (x:_) list = uncurry Q <$> F x list

R. Martinho Fernandes Over a year ago

@FUZxxl Thanks! I knew there was a way to write fmap (uncurry Q) (F x list) that looked elegant without parenthesis. I just couldn't remember what the operator was. I guess my Haskell is getting rusty :(

fuz Over a year ago

@Marthino Fernandes: The problem is, that the Prelude doesn't exports <$> by default.

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