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