On Mon, 4 Aug 2003, Ashley Yakeley wrote: > At 2003-08-04 20:00, Ben Rudiak-Gould wrote: > > >This is a different lambda calculus, with a different beta rule. You can > >see the same effect in the type inference rules for implicit parameters: > >If f has type Int -> String and ?x has type (?x :: Int) => Int, then f ?x > >has type (?x :: Int) => String, i.e. the implicit ?x parameter is lifted > >out of the RHS to become a parameter of the whole application node. This > >rule is what makes implicit parameters implicit. > > Ah. Actually I think the beta rule is unchanged with Haskell's implicit > parameters. Consider: > > x :: (?x :: Int) => Int > f :: Int -> String > f :: ((?x :: Int) => Int) -> ((?x :: Int) => String) -- specialisation > f x :: (?x :: Int) => String -- apply beta rule
I think it amounts to the same thing. This is a nice way of looking at it, though. Either way, the effect is that implicit parameters don't "disappear" into function applications unless the function has been explicitly typed to take them; instead they grow to encompass larger and larger portions of the containing function, until eventually they become parameters of the whole function. Result: you don't have to put them there by hand. > So is this valid or not? > > b :: (?x :: Int) => Int > b = [EMAIL PROTECTED] -> @x > > f :: ((?x :: Int) => Int) -> (Int,Int) > f = \a -> ((a,[EMAIL PROTECTED] -> @x) [EMAIL PROTECTED] = 2}) > > f b :: (Int,Int) > f b = ((b,[EMAIL PROTECTED] -> @x) [EMAIL PROTECTED] = 2}) > = (([EMAIL PROTECTED] -> @x,[EMAIL PROTECTED] -> @x) [EMAIL PROTECTED] = 2}) Yes, it is. This illustrates what you pointed out earlier, that the program's semantics can be changed by adding explicit type signatures which include implicitly-parameterized parameters. > If it is valid, then this must be a valid reduction: > > ((\a -> ((a,[EMAIL PROTECTED] -> @x) [EMAIL PROTECTED] = 2})) ([EMAIL PROTECTED] > -> @x),[EMAIL PROTECTED] -> @x) [EMAIL PROTECTED] = 1} > (([EMAIL PROTECTED] -> @x,[EMAIL PROTECTED] -> @x) [EMAIL PROTECTED] = 2},[EMAIL > PROTECTED] -> @x) [EMAIL PROTECTED] = 1} Not unless you give an explicit type signature, no. Without it the compiler will infer a different type for f which does not have any arguments with implicit parameters, because inferred types never do (see section 2.1 in the original paper). Without this assumption the compiler couldn't infer a principal type. (Specialization doesn't help here, because the set of valid specializations depends on f's internals, and can't be captured by a principal type.) -- Ben _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell