HI
It's one of those things - I know sort of instinctively why it is so
but can't think of the formal rationale for it:
f g x = g (g x) :: (t - t) - (t - t)
Why not
(t - t) - t - (t - t)
to take account of the argument x for g?
Cheers
Paul
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PR Stanley [EMAIL PROTECTED] writes:
It's one of those things - I know sort of instinctively why it is so
but can't think of the formal rationale for it:
f g x = g (g x) :: (t - t) - (t - t)
(t - t) - (t - t)
So
g :: t - t
x :: t
Thus
f :: (t - t) - t - t
(The last parenthesis is
PR Stanley:
I know sort of instinctively why it is so but
can't think of the formal rationale for it:
f g x = g (g x) :: (t - t) - (t - t)
First of all - it is not the definition f g x = ... :: (t- ...
but the type of the function which might be specified:
f :: (t-t)-t-t
Then, the answer
Try putting this through your GHCI:
:t twice f x = f (f x)
I'd presume that based on the inference of (f x) f is (t - t) and x :: t
Yes, Maybe I should get the right associativity rule cleared first.
Cheers,
Paul
At 20:35 01/04/2008, you wrote:
PR Stanley:
I know sort of instinctively why it
