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haskellfunction-composition
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Reputation: 15

Take a list of functions and compose the list with a value

I'm currently attempting to do this problem from the ebook, Haskell School of Music:

Define a function applyAll that, given a list of functions [ f1, f2, ..., fn ] and a value v, returns the result f1 (f2 (...(fn v)...)).

For example: applyAll [simple 2 2, (+3)] 5 ⇒ 20

Currently I have

simple       :: Integer -> Integer -> Integer -> Integer
simple x y z = x * (y + z)

applyAll          :: [(f -> a)] -> a -> a
applyAll [f] v  = foldr (.) v [f]

a = (applyAll [simple 2 2, (+3)] 5)
print a

Which gives me the error:

Couldn't match type `f' with `a0 -> f'
  `f' is a rigid type variable bound by
      the type signature for applyAll :: [f -> a] -> a -> a

Expected type: (f -> a) -> a -> a
  Actual type: (f -> a) -> (a0 -> f) -> a0 -> a
In the first argument of `foldr', namely `(.)'
In the expression: foldr (.) v [f]

I'm assuming it has something to do with the type signature, but nothing I've tried so far has yielded results.

Upvotes: 0

Views: 307

Answers (1)

daniel gratzer
daniel gratzer

Reputation: 53911

The problem here is that the type of . is (b->c) -> (a->b) -> a -> c but our initial value for our fold is not a function. To put this in english, . is function composition, but we want function application. This is easily solved however with ($).

($) f a = f a

so

applyAll :: [(f -> a)] -> a -> a
applyAll [f] v  = foldr ($) v [f]

Next we have a problem with the [f] part. All that does is match a single element list which is probably not what you want.

applyAll fs v = foldr ($) v fs

Now we need to deal with one last part, the type signature. Our function can't take in functions of type f -> a because when we connect them we'll feed a function a value of type a when it expects an f. So the end result is

 applyAll :: [a -> a] -> a -> a
 applyAll fs v = foldr ($) v fs

Upvotes: 3

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