To reiterate what others have said,
readFile . head xs
== readFile . (head xs) { function application binds strongest }
== (.) readFile (head xs) { operators are also functions }
The types,
(.) :: (b -> c) -> (a -> b) -> (a -> c)
readFile :: FilePath -> IO String
head xs :: FilePath
This cannot work as the types don't match. On the other hand, using ($)
instead of (.) will work.
Try writing that out and reasoning with the types on pen and paper, as an
exercise.
If you're interested, there is an excellent post about equational reasoning
here: http://www.haskellforall.com/2013/12/equational-reasoning.html
Enjoy :)
On 4 April 2015 at 08:10, Mike Meyer
As is often the case with Haskell, your answer is in the types:
Prelude> :t ($) ($) :: (a -> b) -> a -> b Prelude> :t (.) (.) :: (b -> c) -> (a -> b) -> a -> c
So $ takes a function and applies it to a value. . takes two functions and composes them and applies the result to a value.
So readFirst xs = (readFile.head) xs, or just readFirst = readFile . head. But readFirst xs = (readFile $ head) xs will also fail, because readFile doesn't work on objects of type head.
On Fri, Apr 3, 2015 at 9:23 PM, Vale Cofer-Shabica < vale.cofershabica@gmail.com> wrote:
Could someone please explain why the commented line fails spectacularly while the final line succeeds?
import System.IO (getContents) import System.Environment (getArgs)
fileInput :: IO String fileInput = getArgs>>=readFirst where readFirst :: [FilePath] -> IO String readFirst [] = System.IO.getContents --readFirst xs = readFile.head xs readFirst xs = readFile $ head xs
I'm particularly confused given the following typings (from ghci):
readFile.head :: [FilePath] -> IO String readFile.head [] :: a -> IO String
And this is still stranger:
:type readFile.head ["foo", "bar"]
<interactive>:28:16: Couldn't match expected type `a0 -> FilePath' with actual type `[Char]' In the expression: "foo" In the first argument of `head', namely `["foo", "bar"]' In the second argument of `(.)', namely `head ["foo", "bar"]'
<interactive>:28:23: Couldn't match expected type `a0 -> FilePath' with actual type `[Char]' In the expression: "bar" In the first argument of `head', namely `["foo", "bar"]' In the second argument of `(.)', namely `head ["foo", "bar"]'
Many thanks in advance, vale _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
On Fri, Apr 3, 2015 at 9:23 PM, Vale Cofer-Shabica < vale.cofershabica@gmail.com> wrote:
Could someone please explain why the commented line fails spectacularly while the final line succeeds?
import System.IO (getContents) import System.Environment (getArgs)
fileInput :: IO String fileInput = getArgs>>=readFirst where readFirst :: [FilePath] -> IO String readFirst [] = System.IO.getContents --readFirst xs = readFile.head xs readFirst xs = readFile $ head xs
I'm particularly confused given the following typings (from ghci):
readFile.head :: [FilePath] -> IO String readFile.head [] :: a -> IO String
And this is still stranger:
:type readFile.head ["foo", "bar"]
<interactive>:28:16: Couldn't match expected type `a0 -> FilePath' with actual type `[Char]' In the expression: "foo" In the first argument of `head', namely `["foo", "bar"]' In the second argument of `(.)', namely `head ["foo", "bar"]'
<interactive>:28:23: Couldn't match expected type `a0 -> FilePath' with actual type `[Char]' In the expression: "bar" In the first argument of `head', namely `["foo", "bar"]' In the second argument of `(.)', namely `head ["foo", "bar"]'
Many thanks in advance, vale _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
-- Regards Sumit Sahrawat