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
PR Stanley
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 not necessary, but implies that the type of the partial application f g is a function t -> t .) -k -- If I haven't seen further, it is by standing in the footprints of giants
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
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 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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
participants (3)
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jerzy.karczmarczuk＠info.unicaen.fr -
Ketil Malde -
PR Stanley