Am I misinterpreting the purpose of fbreak?

[David Waite]

This is what I attempted basically, and it seems to work fine so far
(hooray!)

The only difference is that I had my action check for queued events
before breaking.

-DW

On Jan 5, 5:52 pm, Adrian Thurston <thurs... at cs.queensu.ca> wrote:
> Sorry that won't work. When you break from the processing loop you
> abandon any remaining actions on that character. Ragel does not jump
> back into the action list to the point of departure.
>
> What you could do is queue up actions and then embed the break action using:
>
> ${ p++; break; }
>
> Just be sure the break action is embedded after the actions that queue
> up tokens. See the section on action ordering in the manual.
>
> -Adrian
>
> David Waite wrote:
> > I don't know if advancing p is the correct option all the time though,
> > because (in the example, and unfortunately in the actual machine I'm
> > trying to create) I trigger multiple breaking actions on a single
> > transition. For example, the two character JSON document "[]"
> > currently should have the following actions triggered by my machine:
>
> > main:
> >   StartDocument - fbreak; (returns Event.Document)
> >   ParseArray - fcall array;
> >   StartArray - fbreak; (returns Event.Array)
> >   EndArray - fbreak; (returns Event.EndArray)
> >   Return - fret;
> >   EndDocument (eof) - fbreak; (returns Event.EndDocument)
>
> > If I remove the StartDocument fbreak, would incrementing p on break
> > work?
>
> > -DW
>
> > On Jan 5, 3:56 pm, Adrian Thurston <thurs... at cs.queensu.ca> wrote:
> >> The problem is that fbreak does not advance p when it breaks out of the
> >> loop. This is a choice I made a long time ago and I think I was taking
> >> the "do as little as possible on behalf of the user" approach. But not
> >> advancing p while advancing cs is clearly wrong and confusing. This is
> >> something that will change in 6.0 (it's in the TODO already).
>