Re: [Haskell-beginners] Why this order of parameters

He means why is the first argument a -> b -> b in one function but b -> a -> b in the other. And I have no answer for that. I admit to having been caught by that in the past. You can't just replace one with the other in code, you have to do a flip on the operator. On Thu, Nov 12, 2015 at 12:34 PM, Alex Belanger wrote:

foldr :: (a -> b -> b) -> b -> [a] -> b foldr :: (b -> a -> a) -> a -> [b] -> a

are exactly the same types; simply inverted names for the type variables.

I don't clearly see any benefit.

Alex. On Nov 12, 2015 12:00 PM, "Martin Vlk" wrote:

...

Hi, my first intuition about this is that in data constructor it technically doesn't matter, but you could argue that "a" represents the actual result of the function so it comes first. Second comes the state, which is the side thing, hence the secondary/less important position.

As for the order of type constructor parameters you are right - state is part of the structure that Monoid, Functor, Applicative, Monad and the like use.

Martin

martin:

...

runState :: State s a -> s -> (a, s)

I understand that in the constructor s has to be first, so we can turn (State s) into a monad. But why doesn't s come first in the result too, as in

runState :: State s a -> s -> (s, a)

---

Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners