arity failures Thomas Bushnell BSG (13 Nov 2009 18:54 UTC)
Re: arity failures David Van Horn (13 Nov 2009 19:05 UTC)
Re: arity failures Thomas Bushnell BSG (13 Nov 2009 19:25 UTC)
Re: arity failures Per Bothner (13 Nov 2009 19:16 UTC)
Re: arity failures Thomas Bushnell BSG (13 Nov 2009 19:27 UTC)
Re: arity failures Shiro Kawai (13 Nov 2009 20:13 UTC)

Re: arity failures Thomas Bushnell BSG 13 Nov 2009 19:25 UTC

On Fri, 2009-11-13 at 14:05 -0500, David Van Horn wrote:
> Thomas Bushnell BSG wrote:
> > Also, "failing" an arity check doesn't mean that you will get any kind
> > of error or exception as a result.
> [...]
> > So, to recap:
> >
> > Being told that a procedure takes, say, exactly three arguments, does
> > not mean that
> >   1) You won't get an error for passing three arguments;
> >   2) You will get an error for passing other than three arguments.
> I'm confused -- the intent of the proposal is if you apply a procedure
> to a number of arguments it is reported not to accept, you will get an
> error.  Why is this not the case?

Sorry, there are two ways to go, and I was assuming one branch rather
than spelling out both.  I'm sorry for not being clearer.

A Scheme system may allow secret extra arguments if it wishes.  That is,
it is an error for a programmer to say (define (x a b) ...) and then
call x as (x 1 2 3).  But a system is free to give an implementation
defined meaning to expressions like (x 1 2 3) in that context.

So for a system which does that, there are several choices it could

When it sees (define (x a b) ...) it could *report* that the arity is
indefinite.  That means that programmers cannot predict in a standard
fashion what the arity procedure returns for the lambda expressions they

Or (what I was assuming) it could return an arity object which documents
the syntax of the lambda expression which created the procedure, but
then, an "arity mismatch" would not guarantee an error would be

I was assuming that the second is the choice implementers would choose
(hence my email's wording).  I made that assumption because the only
meaning I can assign to "arity" is the one based upon the syntax of the
formals in the defining lambda expression.