Unary functions and infix notation
As is well known, any binary function f can be turned into an infix operator by surrounding it with backticks: f a b -- prefix application a `f` b -- infix application It is then possible to take left and right sections, i.e. partially applying f: (a `f`) -- equivalent to \b -> a `f` b (`f` b) -- equivalent to \a -> a `f` b This extends relatively naturally to functions of arity greater than two, where usage of a function in infix notation produces a binary operator that returns a function of arity n-2. Weirdly, however, infix notation can also be used for unary functions with polymorphic types, as the following ghci session shows: Prelude> :t (`id` 1) (`id` 1) :: Num a => (a -> t) -> t Prelude> (`id` 1) (\y -> show y ++ ".what") "1.what" Desugaring of an equivalent source file shows that id is applied to the anonymous function, which is then applied to 1. The following example of a function that is not polymorphic in its return type behaves closer to what I would have expected: It does not work. Prelude> let z = (\y -> True) :: a -> Bool Prelude> :t (`z` True) <interactive>:1:2: The operator `z' takes two arguments, but its type `a0 -> Bool' has only one In the expression: (`z` True) What is the purpose/reason for this behaviour? Thank you, --Johannes
On Fri, Sep 6, 2013 at 11:04 AM, Johannes Emerich
Desugaring of an equivalent source file shows that id is applied to the anonymous function, which is then applied to 1.
The following example of a function that is not polymorphic in its return type behaves closer to what I would have expected: It does not work.
Prelude> let z = (\y -> True) :: a -> Bool Prelude> :t (`z` True)
<interactive>:1:2: The operator `z' takes two arguments, but its type `a0 -> Bool' has only one In the expression: (`z` True)
What is the purpose/reason for this behaviour?
Coming from another language, where functions aren't first class, you will probably be used to the notion that a function type is somehow different from any other type. You'll need to unlearn that for functional languages: function types are just as "real" as (Integer) is, and if I have a type variable somewhere which doesn't have constraints otherwise preventing it, that type variable can end up being (Integer) or (a -> a) or (Num c => c -> c -> c) or (Maybe [x]) or (Maybe (a -> a)) or any other (rank-1, i.e. no internal "forall"s) type. (id) has the type (a -> a); in the use mentioned in the first quoted paragraph, this has unified (a) with (b -> b) to produce (id :: (b -> b) -> (b -> b)) in order for the whole expression to be typeable. In your second example, you don't have polymorphism "where it's needed" so it can't infer a type that will work. -- brandon s allbery kf8nh sine nomine associates allbery.b@gmail.com ballbery@sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net
The observation that this only applies to functions with a polymorphic
return type is key.
id :: a -> a
This can be instantiated at
id' :: (a->b) -> (a->b)
id' :: (a->b) -> a -> b -- these are the same
What this means is that id is a function with arity-2 whenever the first
argument is arity-1, and generally id is a function of arity x+1 where x is
the argument arity. Incidentally, this is exactly the same as the ($)
operator.
John L.
On Fri, Sep 6, 2013 at 10:04 AM, Johannes Emerich
As is well known, any binary function f can be turned into an infix operator by surrounding it with backticks:
f a b -- prefix application a `f` b -- infix application
It is then possible to take left and right sections, i.e. partially applying f:
(a `f`) -- equivalent to \b -> a `f` b (`f` b) -- equivalent to \a -> a `f` b
This extends relatively naturally to functions of arity greater than two, where usage of a function in infix notation produces a binary operator that returns a function of arity n-2.
Weirdly, however, infix notation can also be used for unary functions with polymorphic types, as the following ghci session shows:
Prelude> :t (`id` 1) (`id` 1) :: Num a => (a -> t) -> t Prelude> (`id` 1) (\y -> show y ++ ".what") "1.what"
Desugaring of an equivalent source file shows that id is applied to the anonymous function, which is then applied to 1.
The following example of a function that is not polymorphic in its return type behaves closer to what I would have expected: It does not work.
Prelude> let z = (\y -> True) :: a -> Bool Prelude> :t (`z` True)
<interactive>:1:2: The operator `z' takes two arguments, but its type `a0 -> Bool' has only one In the expression: (`z` True)
What is the purpose/reason for this behaviour?
Thank you, --Johannes _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
On Fri, Sep 06, 2013 at 05:04:12PM +0200, Johannes Emerich wrote:
Weirdly, however, infix notation can also be used for unary functions with polymorphic types, as the following ghci session shows:
Prelude> :t (`id` 1) (`id` 1) :: Num a => (a -> t) -> t Prelude> (`id` 1) (\y -> show y ++ ".what") "1.what"
There's nothing special about infix notation here: Prelude> :t \x -> id x 1 \x -> id x 1 :: Num a => (a -> t) -> t Prelude> (\x -> id x 1) (\y -> show y ++ ".what") "1.what" Tom
But we can do next: Prelude> :set XPostfixOperators Prelude> let z = (\y -> True) :: a -> Bool Prelude> :t (True `z`) But still `z` True ~ \a -> a `z` True ~ \a -> z a True and `z` must be a function with minimum 2 arguments -- View this message in context: http://haskell.1045720.n5.nabble.com/Unary-functions-and-infix-notation-tp57... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.
participants (5)
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Brandon Allbery -
Johannes Emerich -
John Lato -
Tom Ellis -
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