CODEWITHSUNDEEP
haskellfunction-compositionpointfree
user2237264
user2237264

Reputation:

Application of arguments to function composition in Haskell

Being a newbie to Haskell I can’t understand why the expression head . words “one two three four” throws an exception and function composition head . words must be applied with $ operator - the expression on the right of it doesn’t need further evaluation because it’s just a single String. The other way to compile it is to put head . words in parentheses but (head . words) :: String -> String has the same type as head . words :: String -> String so why putting it in parentheses makes the expression compile?

Upvotes: 6

Views: 297

Answers (1)

Will Ness
Will Ness

Reputation: 71070

Because of precedence rules. Application has highest precedence; $ - lowest.

head . words “one two three four” is parsed as head . (words “one two three four”) i.e. words applied on a string must produce a function (as demanded by (.)). But that's not the type that words has:

Prelude> :t words
words :: String -> [String]

head . words $ “one two three four” on the other hand, is parsed as (head . words) “one two three four” and the types fit.

Upvotes: 11

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