CODEWITHSUNDEEP
haskell
Bercovici Adrian
Bercovici Adrian

Reputation: 9360

Dealing with null method argument in Haskell

I have tried implementing my custom map function and i get this error which does not tell me much.

mymap::(a->b)->[a]->[b]
mymap _ [] =[]
mymap null (x:xs)=x:xs
mymap f (x:xs)=f x : mymap f xs

Error message:

* Couldn't match type `a' with `b'
      `a' is a rigid type variable bound by
        the type signature for:
          mymap :: forall a b. (a -> b) -> [a] -> [b]

Why is it not correct since i provide a predicate from a to b, a source list of a and i expect a list of b's

Upvotes: 0

Views: 65

Answers (1)

AJF
AJF

Reputation: 11923

There is no null pointer in Haskell, so this line:

mymap null (x:xs) = (x:xs)

is equivalent to

mymap _ s = s

Since null simply matches everything regardless of value, being a valid identifier. You could say, for instance, null = 6, and it would be valid in Haskell.

Thus giving the function type mymap :: b -> [a] -> [a], which is different from the type you intended.

You should remove that line. The 'correct' implementation is:

mymap :: (a -> b) -> [a] -> [b]
mymap _ []     = []
mymap f (x:xs) = f x : map f xs

You're right to pattern match on both [] and (:), because in the declaration of the list type, both occur:

-- Compiler magic occurs here! This is not usually valid syntax
data [a] = [] | a : [a]

We therefore see that in a list there are only two possibilities, both of which we must match. However, this is not the case with functions.

Upvotes: 3

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