:: (a -> b) -> a -> b -package:base

Delay inlining a function until late in the game (simplifier phase 0).

Application operator. This operator is redundant, since ordinary application (f x) means the same as (f $ x). However, $ has low, right-associative binding precedence, so it sometimes allows parentheses to be omitted; for example:

f $ g $ h x  =  f (g (h x))

It is also useful in higher-order situations, such as map ($ 0) xs, or zipWith ($) fs xs. Note that ($) is representation-polymorphic in its result type, so that foo $ True where foo :: Bool -> Int# is well-typed.

Strict (call-by-value) application operator. It takes a function and an argument, evaluates the argument to weak head normal form (WHNF), then calls the function with that value.

($) is the function application operator. Applying ($) to a function f and an argument x gives the same result as applying f to x directly. The definition is akin to this:

($) :: (a -> b) -> a -> b
($) f x = f x

This is id specialized from a -> a to (a -> b) -> (a -> b) which by the associativity of (->) is the same as (a -> b) -> a -> b. On the face of it, this may appear pointless! But it's actually one of the most useful and important operators in Haskell. The order of operations is very different between ($) and normal function application. Normal function application has precedence 10 - higher than any operator - and associates to the left. So these two definitions are equivalent:

expr = min 5 1 + 5
expr = ((min 5) 1) + 5

($) has precedence 0 (the lowest) and associates to the right, so these are equivalent:

expr = min 5 $ 1 + 5
expr = (min 5) (1 + 5)

Examples

A common use cases of ($) is to avoid parentheses in complex expressions. For example, instead of using nested parentheses in the following Haskell function:

-- | Sum numbers in a string: strSum "100  5 -7" == 98
strSum :: String -> Int
strSum s = sum (mapMaybe readMaybe (words s))

we can deploy the function application operator:

-- | Sum numbers in a string: strSum "100  5 -7" == 98
strSum :: String -> Int
strSum s = sum $ mapMaybe readMaybe $ words s

($) is also used as a section (a partially applied operator), in order to indicate that we wish to apply some yet-unspecified function to a given value. For example, to apply the argument 5 to a list of functions:

applyFive :: [Int]
applyFive = map ($ 5) [(+1), (2^)]
>>> [6, 32]

Technical Remark (Representation Polymorphism)

($) is fully representation-polymorphic. This allows it to also be used with arguments of unlifted and even unboxed kinds, such as unboxed integers:

fastMod :: Int -> Int -> Int
fastMod (I# x) (I# m) = I# $ remInt# x m

Strict (call-by-value) application operator. It takes a function and an argument, evaluates the argument to weak head normal form (WHNF), then calls the function with that value.

Application operator. This operator is redundant, since ordinary application (f x) means the same as (f $ x). However, $ has low, right-associative binding precedence, so it sometimes allows parentheses to be omitted; for example:

f $ g $ h x  =  f (g (h x))

It is also useful in higher-order situations, such as map ($ 0) xs, or zipWith ($) fs xs. Note that ($) is levity-polymorphic in its result type, so that foo $ True where foo :: Bool -> Int# is well-typed.

Application operator. This operator is redundant, since ordinary application (f x) means the same as (f $ x). However, $ has low, right-associative binding precedence, so it sometimes allows parentheses to be omitted; for example:

f $ g $ h x  =  f (g (h x))

It is also useful in higher-order situations, such as map ($ 0) xs, or zipWith ($) fs xs.

In Haskell apply the first argument to the second argument, in HDL just return the second argument. This is used in the generated pack/unpack to not do anything in HDL.

Strict (call-by-value) application operator. It takes a function and an argument, evaluates the argument to weak head normal form (WHNF), then calls the function with that value.

Application operator. This operator is redundant, since ordinary application (f x) means the same as (f $ x). However, $ has low, right-associative binding precedence, so it sometimes allows parentheses to be omitted; for example:

f $ g $ h x  =  f (g (h x))

It is also useful in higher-order situations, such as map ($ 0) xs, or zipWith ($) fs xs. Note that ($) is levity-polymorphic in its result type, so that foo $ True where foo :: Bool -> Int# is well-typed

Right-associative apply operator. Read as "apply backward" or "pipe from". Use this to create long chains of computation that suggest which direction things move in. You may prefer this operator over (|>) for IO actions since it puts the last function first.

>>> print <| negate <| recip <| succ <| 3
-0.25

Or use it anywhere you would use ($). Note that (<|) and (|>) have the same precedence, so they cannot be used together.

>>> -- This doesn't work!

>>> -- print <| 3 |> succ |> recip |> negate

\ x -> (f <| x) == f x

\ x -> (g <| f <| x) == g (f x)

Right-associative apply' operator. Read as "strict apply backward" or "strict pipe from". Use this to create long chains of computation that suggest which direction things move in. You may prefer this operator over (!>) for IO actions since it puts the last function first.

>>> print <! negate <! recip <! succ <! 3
-0.25

Or use it anywhere you would use ($!). The difference between this and (<|) is that this evaluates its argument before passing it to the function.

>>> const True <| undefined
True

>>> const True <! undefined
*** Exception: Prelude.undefined
...

Note that (<!) and (!>) have the same precedence, so they cannot be used together.

>>> -- This doesn't work!

>>> -- print <! 3 !> succ !> recip !> negate

\ x -> (f <! x) == seq x (f x)

\ x -> (g <! f <! x) == let y = seq x (f x) in seq y (g y)

($) is the function application operator. Applying ($) to a function f and an argument x gives the same result as applying f to x directly. The definition is akin to this:

($) :: (a -> b) -> a -> b
($) f x = f x

This is id specialized from a -> a to (a -> b) -> (a -> b) which by the associativity of (->) is the same as (a -> b) -> a -> b. On the face of it, this may appear pointless! But it's actually one of the most useful and important operators in Haskell. The order of operations is very different between ($) and normal function application. Normal function application has precedence 10 - higher than any operator - and associates to the left. So these two definitions are equivalent:

expr = min 5 1 + 5
expr = ((min 5) 1) + 5

($) has precedence 0 (the lowest) and associates to the right, so these are equivalent:

expr = min 5 $ 1 + 5
expr = (min 5) (1 + 5)

Examples

A common use cases of ($) is to avoid parentheses in complex expressions. For example, instead of using nested parentheses in the following Haskell function:

-- | Sum numbers in a string: strSum "100  5 -7" == 98
strSum :: String -> Int
strSum s = sum (mapMaybe readMaybe (words s))

we can deploy the function application operator:

-- | Sum numbers in a string: strSum "100  5 -7" == 98
strSum :: String -> Int
strSum s = sum $ mapMaybe readMaybe $ words s

($) is also used as a section (a partially applied operator), in order to indicate that we wish to apply some yet-unspecified function to a given value. For example, to apply the argument 5 to a list of functions:

applyFive :: [Int]
applyFive = map ($ 5) [(+1), (2^)]
>>> [6, 32]

Technical Remark (Representation Polymorphism)

($) is fully representation-polymorphic. This allows it to also be used with arguments of unlifted and even unboxed kinds, such as unboxed integers:

fastMod :: Int -> Int -> Int
fastMod (I# x) (I# m) = I# $ remInt# x m

This operator is an alternative for $ with a higher precedence. It is suitable for usage within applicative style code without the need to add parenthesis.

Deprecated: use % instead

Memoize a unary function.

Memoize a unary function.