[racket] Typed racket puzzle (ii)

From: Norman Gray (norman at astro.gla.ac.uk)
Date: Wed Aug 17 09:42:48 EDT 2011

Sam, hello.

On 2011 Aug 17, at 13:12, Sam Tobin-Hochstadt wrote:

>> OoooKay.  I was thinking of subtyping in terms of a naive mixture of the (set) extensions of types (the set of things which are instances of (List String String) is a subset of the things which are instances of (Listof String)), and assignment-compatibility.  That understanding clearly fails in your example, because while (List String String) can be assigned to a (Listof String), the function test is declared to accept -- in the sense of have bound to its arguments -- (Listof String) which in this case isn't true.  Gotcha (I think).
> It's ok to think of things as sets, and subtyping as subsets.  The
> trick is that functions are a little different than you expect. In
> particular, if you have two types, (A -> B) and (C -> D), for the
> first to be a subtype of the second, you do need B to be a subtype of
> D.  But for the arguments, it goes the other way: C must be a subtype
> of A.

Thanks for the explanation.  Curses!  I must have been reasonably close to working that out for myself, with the thought about what the function is intended to accept.

> Here's how to see why you need that. To say that (A -> B) is a subtype
> of (C -> D), you're saying that you can use (A -> B) anywhere you need
> (C -> D).  But some program that expects a (C -> D) function might
> pass it a C, of course, so your new replacement function must accept
> all of the Cs.  Therefore, A needs to accept all of the Cs, which
> means that A must be a subtype of C.

Don't you mean "C must be a subtype of A", here?

> The standard term for this is that function types are "contravariant"
> (go the other way) in their argument type.

Thanks for 'contravariant' in this context.  It's led me to the relevant application of category theory, which I think is the underlying structure I was looking for.  I suppose it's effectively a combination of 'define' and ':' which is in practice the functor in TR terms, with subtyping being the morphism in the type category.  Or something like that.

All is now clear.

Best wishes,


Norman Gray  :  http://nxg.me.uk
School of Physics and Astronomy, University of Glasgow, UK

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