From: Brian Mitchell <binary42@...>
Date: 2006-09-26T08:48:35+09:00
Subject: Re: Riddle me this (a question about expressions)

On 9/25/06, Rick DeNatale <rick.denatale@gmail.com> wrote:
> On 9/25/06, Brian Mitchell <binary42@gmail.com> wrote:
> > On 9/25/06, Rick DeNatale <rick.denatale@gmail.com> wrote:
> >> I think that the proper documentation of
> > > Kernel#raise/Thread#raise should be that it causes a non-local return
> > > unless it is rescued, in which case the value is the value of the
> > > expression in the rescue.  That's how it works now.
> >
> > Well, those words are sort of confusing:
> >
> > begin
> >   @x = raise
> > rescue
> >   42
> > end
> >
> > p @x #=> nil
> >
> > @x = raise rescue 42
> >
> > p @x #=> 42
>
> No it's the same thing.  The value of the lvalue in this case is the
> value of raise, which since the raise was rescued is 42.
>
> I think that my proposed documentation covers this case.
>

No it doesn't. I would say that is was rescued in my example as well.
The language might look like:

Kernel#raise/Thread#raise causes a non-local return unless it is
rescued, in which case the result _of the begin..end block_ is the
result of the expression in the rescue clause.

> > To put the result into better perspective, the binding of = is to the
> > entire expression including the rescue, which has the result of 42
> > [1].
>
> Except in this case the only 'entire expression' which includes the rescue is:
>
> begin; x=raise;rescue; 42;end
>
> And x is not being assigned the value of this expression.
>

Actually, single line rescue has tighter bindings than =. So it would
be like I show.

x = raise rescue 42

becomes:

 x = (raise rescue 42)

> >This makes much more sense as one could imagine an invisble
> > wrapper like:
> >
> > @x = begin raise rescue 42 end
>
> But I think it makes even more sense to explain the syntax as it is
> rather than modifying it.
>

That was my attempt. I am explaining that:

x = raise "something" rescue 42

has the same result as:

x = begin raise "something" rescue 42 end

This dispelling any doubt about what other differences might be. We
could safely use such a mechanism mentally if it happened to help us.

> > So should raise w/o the rescue be illegal? (no begin/end semantically)
>
> Too fussy, I think.
>

I'm not sure what you mean here. I am trying to say that it is overly
complex to think of raise as an expression which returns. It's kind of
like trying to get the result of calling a continuation; it doesn't
return one. It is much easier to use a structured interpretation where
there are begin..end clauses which happen to behave very dependably.

Brian.