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] [EMAIL PROTECTED]
system administrator [openafs,heimdal,too many hats] [EMAIL PROTECTED]
electrical and computer engineering, carnegie mellon university KF8NH
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