>This is pretty much how it works internally, although the precise
>mechanisms used are slightly different in the bytecode interpreter and
>in the native-code compiler.  Viewed from the outside, it maintains
>the illusion that every function takes only one parameter,
>and fun x y -> ... behaves like fun x -> (fun y -> ...).

Okay, that makes sense, thanks.  Now my question is, is there any way to introspect to get at the arity information?  I think it must be stored somewhere at runtime, because I can pass both f's to another function because they have the same "illusory type".  

Or do they?  Is int -> int -> int typechecked _exactly_ like int -> (int -> int)?  I assume they must be or the illusion wouldn't hold up.  I'd also assume this arity check only happens during application and doesn't affect typechecking at all.

Chris


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