On Dec 27, 2006, at 22:55 , michael rice wrote:
By similar reasoning the always function would seem to have a signature
a -> (b -> a)
where the first argument is just a value and the return value is a function that when given a possibly different value just returns the value originally given to always?
Is that reasoning OK? Are
a -> (b -> a) and a -> b -> a the same signature?
This is a point that has been glossed over a bit: Haskell has the notion of partial application. If you want to start with a function that takes two values, and return a function that takes one value and uses the one previously passed in, you just invoke the function with one parameter; Haskell will produce a function which takes a single argument to complete the expression. Using (*) (prefix version of multiplication) as an example: Prelude> :t ((*) 2) ((*) 2) :: (Num t) => t -> t Prelude> let x2 = ((*) 2) in x2 5 10 This shows the equivalence of the type signatures (a -> a -> a) and (a -> (a -> a)), and is one of the strengths of Haskell: you can pass a section (a "partially expanded" function") wherever a function is expected. Prelude> map ((*) 2) [1..5] [2,4,6,8,10] This doesn't only work for prefix functions, by the way; the above example is more naturally written as (2*): Prelude> :t (2*) (2*) :: (Num t) => t -> t Prelude> map (2*) [1..5] [2,4,6,8,10] You can also say (*2), which provides the right-hand argument; this is useful for non-commutative functions like (/). But don't try it with (-), because you'll trip over an unfortunate parsing hack for negative numbers: Prelude> :t (-2) -- whoops, it's a number, not a function! (-2) :: (Num a) => a The Prelude provides a workaround for this, though: Prelude> :t (subtract 2) (subtract 2) :: (Num t) => t -> t -- brandon s. allbery [linux,solaris,freebsd,perl] allbery@kf8nh.com system administrator [openafs,heimdal,too many hats] allbery@ece.cmu.edu electrical and computer engineering, carnegie mellon university KF8NH