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type (f :: Type -> Type)
$ a = f adata-effects-core Control.Effect Type-level infix applcation for functors.
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mixed-types-num Numeric.MixedTypes.PreludeHiding ($) is the function application operator. Applying ($) 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
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LambdaHack Game.LambdaHack.Core.Prelude ($) is the function application operator. Applying ($) 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
-
cabal-install-solver Distribution.Solver.Compat.Prelude ($) is the function application operator. Applying ($) 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
type (f :: Type -> Type)
$ a = f aheftia Control.Monad.Hefty Type-level infix applcation for functors.
type (f :: Type -> Type)
$ a = f aheftia Control.Monad.Hefty Type-level infix applcation for functors.
-
ihaskell IHaskellPrelude ($) is the function application operator. Applying ($) 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
-
incipit-base Incipit.Base ($) is the function application operator. Applying ($) 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
-
clash-prelude Clash.HaskellPrelude ($) is the function application operator. Applying ($) 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
-
dimensional Numeric.Units.Dimensional.Prelude ($) is the function application operator. Applying ($) 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