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I wrote:
...Will (id :: A -> A $!) do the trick?
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.
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 -Yitz
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