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:[EMAIL PROTECTED]>[EMAIL PROTECTED]> 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
_______________________________________________
Haskell-Cafe mailing list
[email protected]
http://www.haskell.org/mailman/listinfo/haskell-cafe
