On Fri, Feb 13, 2009 at 05:21:50PM +0100, Thomas Davie wrote:
Hey,
Thanks for all the suggestions. I was hoping that there was some
uniform
pattern that would extend to n arguments (rather than having to use
liftM2, litM3, etc. or have different 'application' operators in
between
Hi Conor,
Will this do?
http://www.haskell.org/haskellwiki/Idiom_brackets
You get to write
iI f a1 a2 a3 Ji
for
do x1 - a1
x2 - a2
x3 - a3
f a1 a2 a3
amongst other things...
Cool :-) I had seen those idiom brackets before and put them on my
mental
Hey,
Thanks for all the suggestions. I was hoping that there was some uniform
pattern that would extend to n arguments (rather than having to use
liftM2, litM3, etc. or have different 'application' operators in between
the different arguments); perhaps not. Oh well :)
Thanks again!
Edsko
Hi Edsko
On 13 Feb 2009, at 09:14, Edsko de Vries wrote:
Hey,
Thanks for all the suggestions. I was hoping that there was some
uniform
pattern that would extend to n arguments (rather than having to use
liftM2, litM3, etc. or have different 'application' operators in
between
the
Hey,
Thanks for all the suggestions. I was hoping that there was some
uniform
pattern that would extend to n arguments (rather than having to use
liftM2, litM3, etc. or have different 'application' operators in
between
the different arguments); perhaps not. Oh well :)
Sure you can!
Check out liftM2. It's almost what you want.
On Thu, Feb 12, 2009 at 6:36 PM, Edsko de Vries devri...@cs.tcd.ie wrote:
Hi,
I can desugar
do x' - x
f x'
as
x = \x - f x'
which is clearly the same as
x = f
However, now consider
do x' - x
y' - y
f x' y'
Hello,
You could do:
(f = x) = y
?
- jeremy
At Thu, 12 Feb 2009 23:36:19 +,
Edsko de Vries wrote:
Hi,
I can desugar
do x' - x
f x'
as
x = \x - f x'
which is clearly the same as
x = f
However, now consider
do x' - x
y' - y
f x' y'
On Thu, 2009-02-12 at 23:36 +, Edsko de Vries wrote:
Hi,
I can desugar
do x' - x
f x'
as
x = \x - f x'
which is clearly the same as
x = f
However, now consider
do x' - x
y' - y
f x' y'
desugared, this is
x = \x - y = \y' - f x' y'
I
oops, I take that back. It only appears to work if you are sloppy:
x :: (Monad m) = m a
x = undefined
y :: (Monad m) = m b
y = undefined
g :: (Monad m) = a - b - m c
g = undefined
ex1 :: (Monad m) :: m c
ex1 = (g = x) = y
But, if you try to pin down the types you find it only works because
