Partial application: Difference between revisions
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In this example, <hask>addOne</hask> is the result of partially applying <hask>add</hask>. It is a new function that takes | In this example, <hask>addOne</hask> is the result of partially applying <hask>add</hask>. It is a new function that takes an integer, adds 1 to it and returns that as the result. | ||
The important property here is that the <hask>-></hask> operator is right associative, and function application is left associative, meaning the type signature of add actually looks like this: | The important property here is that the <hask>-></hask> operator is right associative, and function application is left associative, meaning the type signature of add actually looks like this: |
Revision as of 22:22, 24 July 2007
Partial application in Haskell involves passing less than the full number of arguments to a function that takes multiple arguments.
Examples
Simple introduction
For example:
add :: Int -> Int -> Int
add x y = x + y
addOne = add 1
In this example, addOne
is the result of partially applying add
. It is a new function that takes an integer, adds 1 to it and returns that as the result.
The important property here is that the ->
operator is right associative, and function application is left associative, meaning the type signature of add actually looks like this:
add :: Int -> (Int -> Int)
This means that add actually takes one argument and returns a function that takes another argument and returns an Int.
Slightly more complicated
What if you have a higher order function?
For example:
comp2 :: (a -> b) -> (b -> b -> c) -> (a -> a -> c)
comp2 f g = (\x y -> g (f x) (f y))
Remembering the maxim that: Functions are not partial, you can partially apply a function.
So, is this a partial definition of comp2'
?
comp2' f = (\x y -> add (f x) (f y))
Not really, this is the definition of another function. But you can achieve the desired result by partially applying comp2
, like so:
comp2' f = comp2 f add