CODEWITHSUNDEEP
haskellfold
newnewbie
newnewbie

Reputation: 1003

Very basic task using foldr

I just used a very simple example used in some lecture notes with my ghci:

 foldr (:) [] 1 2

expecting the result

   [1,2]

However, I get an error. I get an error everytime when I try to use ++ or : as the function given to foldr.

Apparently I am making some pretty obvious mistake but I still cannot seem to find it.

Can anyone help?

Upvotes: 0

Views: 213

Answers (1)

firefrorefiddle
firefrorefiddle

Reputation: 3805

You used foldr like a variadic function by passing it two arguments 1 and 2 instead of [1, 2].

When you run into trouble like that, just check the function's type. You can do that in GHCi:

Prelude> :t foldr
foldr :: (a -> b -> b) -> b -> [a] -> b

So you see that the first argument should be a function (a -> b -> b). You used (:) for that, which is o.k. You can check the type of the partially applied function as well:

Prelude> :t (:)
(:) :: a -> [a] -> [a]

Substituting b with [a] gives us: 

Prelude> :t foldr (:)
foldr (:) :: [a] -> [a] -> [a]

Next, you gave [] as a base case.

Prelude> :t foldr (:) []
foldr (:) [] :: [a] -> [a]

So the resulting function is of type [a] -> [a]. What should you make of this? You have to pass it a list to get a list back! Passing the argument list [1, 2]:

Prelude> :t foldr (:) [] [1,2]
foldr (:) [] [1,2] :: Num a => [a]

Is accepted by the type checker and yields the result:

Prelude> foldr (:) [] [1,2]
[1,2]

I hope this helps type-debugging your programs...

Upvotes: 14

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