无法将预期类型“a”与实际类型“[a]”匹配

Question

I have this list in which I want to go through and convert certain elements of the list into a string. The first 3 functions work, however, the last one (placeRainfull) does not. When I try to load the script, I get this error:

Couldn't match expected type ‘Place’ with actual type ‘[Place]’

I want the function to go through each element in the list and have it run through the addDayWithRainfull function.

Code

getRainfull :: Place -> (String, [Int])
getRainfull (Place p _ _ rf ) = (p, rf)

convrtIntArray :: [Int] -> [String]
convrtIntArray rainfullArray = map show [ i | i <- rainfullArray]

addDayWithRainfull :: (String, [Int]) -> String
addDayWithRainfull (p, rf) = p ++ " " ++ unwords (convrtIntArray rf)

placeRainful :: [Place] -> String
placeRainful places = addDayWithRainfull (getRainfull places)

Answer 1

You've told the compiler that placeRainful takes a list of Place s ( [Place] ), but then you're passing that list of places to getRainFull , which you have indicated takes only a single Place .

In other words, Haskell is inferring based on the implementation of placeRainful that it should have type Place -> String , but then you said that that's not it's type.

As a side note, based on this several similar questions you've asked, it seems like you're having trouble differentiating the types [Place] and Place . Haskell (and indeed, most programming languages) cares a great deal about this distinction. Think about the data that each type represent:

[1, 2, 3, 4, 5] :: [Int]

1 :: Int

These types need to be treated differently because it's very clear what a should be in this expression:

let a = 1 * 1

but it's not at all clear what it should be in this expression:

let a = [1, 2, 3] *  [1, 2, 3]

Answer 2

The problem is that you defined a function:

getRainfull ::  -> (String, [Int])

which takes as input a single Place object, but in your placeRainful you write:

placeRainful :: [Place] -> String
placeRainful  = addDayWithRainfull (getRainfull )

so here you expect to process a list of Place s, that will not work. You can for example make use of concatMap:: Foldable f => (a -> [b]) -> fa -> [b] to process each Place in the list of places and then concatenate these together:

placeRainful :: [Place] -> String
placeRainful places =  (addDayWithRainfull . getRainfull) places

or with [a] -> [[a]] -> [a] :

import Data.List(intercalate)

placeRainful :: [Place] -> String
placeRainful places =  (map (addDayWithRainfull . getRainfull) places)

or perhaps you need to use some other function, but the general idea is that you will need something to pass elements of the list to getRainfull , not the entire list itself.