19 Jan
2007
19 Jan
'07
5:23 a.m.
On Jan 18, 2007, at 10:09 PM, Yitzchak Gale wrote:
I wrote:
Will (id :: A -> A $!) do the trick?
Ulf Norell wrote:
The problem is not with id, it's with composition. For any f and g we have
f . g = \x -> f (g x)
So _|_ . g = \x -> _|_ for any g.
OK, so then how about
f .! g = ((.) $! f) $! g
That should probably do the trick. / Ulf