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 to:
Why not
(t -> t) -> t -> (t -> t)
to take account of the argument x for g?
is simple. If t is the type of x, then g must be g :: t->t, you're right.
So f :: (t->t) -> t -> [the type of the result]
But this result is of the type t, it is g(g x), not (t->t), it is as
simple as that. Perhaps you didn't recognize that "->" is syntactically
a right-associative op, so
a->b->c is equivalent to a->(b->c), or
(t->t)->t->t equiv. to (t->t)->(t->t)
Jerzy Karczmarczuk
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