
2 Apr
2008
2 Apr
'08
4:13 p.m.
2008/4/2, Dan Piponi
On Tue, Apr 1, 2008 at 2:07 PM, PR Stanley
wrote: All you'd have to do is to give the inner most function the highest precdence
What's the innermost function in "f g x" here?
test :: (a -> b -> c) -> a -> b -> c test f g x = f g x
g (if I followed correctly). But we have a problem: If I want f g x to be parsed as f (g x), I would probably want, f g x y to be parsed as f (g x y), considering g as the innermost function, again. I'm almost certain that it would render type inference impossible, and maybe partial application as well. (id 4 is a total application, but id (\x -> x+1) is a partial one...) The only choices left are right associativity or left associativity. Loup