[Haskell-beginners] Understanding the type signature of flip$id

Jeff Lasslett jeff.lasslett at gmail.com
Wed Mar 2 00:43:23 CET 2011


I have been reading about how to use system.console.getOpt,
specifically "Interpreting flags as transformations of an options
record" which can be found at

Line 3 of

compilerOpts argv =
       case getOpt Permute options argv of
          (o,n,[]  ) -> return (foldl (flip id) defaultOptions o, n)
          (_,_,errs) -> ioError (userError (concat errs ++ usageInfo
header options))
      where header = "Usage: ic [OPTION...] files..."

uses (flip id) to resolve a list of functions down to a single record.

Looking at the type of flip$id I can see why it works in this context.
 I just don't understand how passing id to flip results in the
following type signature:

Prelude> :t flip$id
flip$id :: b -> (b -> c) -> c

I do understand what flip has done to map here:-

Prelude> :t flip$map
flip$map :: [a] -> (a -> b) -> [b]

map take a function and a list and produces a new list.  If map is
passed to flip the result is a function that takes a list, then a
function and results in a new list.

How do we go from flip having this signature:

Prelude> :t flip
flip :: (a -> b -> c) -> b -> a -> c

and id having

Prelude> :t id
id :: a -> a

to flip$id looking like flip$id :: b -> (b -> c) -> c   ???


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