What do you need that for?
Can you live with
infixl |$|
(|$|) :: [a - r] - a - [r]
fs |$| x = map ($ x) fs
and, instead of broadcast fs a b use
fs |$| a |$| b
?
On 13 Feb 2009, at 02:34, John Ky wrote:
Hi Haskell Cafe,
I tried using type families over functions, but when I try it
On Fri, 2009-02-13 at 10:34 +1100, John Ky wrote:
Hi Haskell Cafe,
I tried using type families over functions, but when I try it
complains that the two lines marked conflict with each other.
class Broadcast a where
type Return a
broadcast :: a - Return a
instance Broadcast [a -
Hi Miguel,
That's a nice way of writing it.
Thanks,
-John
On Fri, Feb 13, 2009 at 10:42 AM, Miguel Mitrofanov
miguelim...@yandex.ruwrote:
What do you need that for?
Can you live with
infixl |$|
(|$|) :: [a - r] - a - [r]
fs |$| x = map ($ x) fs
and, instead of broadcast fs a b use
Hi Johnaton,
Ah yes. That makes sense. Is there a way to define type r to be all types
except functions?
-John
On Fri, Feb 13, 2009 at 10:44 AM, Jonathan Cast
jonathancc...@fastmail.fmwrote:
On Fri, 2009-02-13 at 10:34 +1100, John Ky wrote:
Hi Haskell Cafe,
I tried using type families
On Fri, 2009-02-13 at 11:15 +1100, John Ky wrote:
Hi Johnaton,
Ah yes. That makes sense. Is there a way to define type r to be all
types except functions?
Not without overlapping instances. I *think* if you turn on {-#
LANGUAGE OverlappingInstances #-} then
instance Broadcast r where
Can you live with
infixl |$|
(|$|) :: [a - r] - a - [r]
fs |$| x = map ($ x) fs
and, instead of broadcast fs a b use
fs |$| a |$| b
?
map ($ x) fs
= { Applicative Functors satisfy... }
pure ($ x) * fs
= { 'interchange' rule from Control.Applicative }
fs * pure x
Thus;
