Interesting. I've never seen it proposed as right-associative before. Just
FYI, the implementation is as simple as this:
infixr 1 |^
(|^) :: a -> (a -> b) -> b
x |^ f = f x
Then you can write:
3 |^ 2 |^ (^) -- produces 2^3 = 8
It seems very odd to me. I don't know why you'd want to apply the arguments
backwards one by one. However, lens and diagrams both provide examples
where you want to start with a value and apply functions "forwards" one by
one, which is why their corresponding operators are left-associative.
-- Dan Burton
On Wed, Apr 16, 2014 at 4:49 PM, Alexey Muranov
On Thursday, April 17, 2014 1:38:48 AM UTC+2, Richard A. O'Keefe wrote:
On 17/04/2014, at 10:25 AM, Alexey Muranov wrote:
For whatever it is worth, i would like to propose for discussion a syntax for "(flip ($))" operation in Haskell.
Oh, you mean like F#'s "|>" operator?
If we were to copy an operator from F#, it would have been nice to copy the F# name for it. Sadly, Data.Sequence already uses |> . Hoogle doesn't find a (|^), so that might work.
No, in fact i wanted to copy it from the exponential notation for function application. In particular, it would need to be right-associative:
y |^ x |^ f == y |^ (x |^ f) == f x y
Maybe a left-associative version can be defined too, by something like this:
x ^| g ^| f == (x ^| (g |^ f) = f (g x)
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