[Haskell-cafe] Unary functions and infix notation
John Lato
jwlato at gmail.com
Fri Sep 6 17:43:01 CEST 2013
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 <johannes at emerich.de>wrote:
> 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
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