23 Oct
2007
23 Oct
'07
2:41 a.m.
f x = x x :: a f x :: b therefore f :: a -> b x = a and x = b therefore a = b therefore f :: a -> a Simple mappings are easy to work out. It's the more detailed stuff I'm not sure about. f g x y = g x (y x) Cheers, Paul At 03:15 23/10/2007, you wrote:
On 10/22/07, PR Stanley <mailto:prstanley@ntlworld.comprstanley@ntlworld.com> wrote: Hi What are the rules for calculating function types? Is there a set procedure ? Thanks, Paul
There must be a set procedure, since otherwise the compiler could not function! =)
Seriously, though, I'm not exactly sure what you're asking for. Could you maybe provide a few examples of the kind of thing you're asking about?
-Brent