On Sun, Apr 10, 2005 at 12:02:35AM +0200, Andre Poenitz wrote: > On Sat, Apr 09, 2005 at 05:33:18PM +0300, Martin Vermeer wrote: > > Here is the next patch... adding a cur.undispatched() statement did > > the job. So the LFUN_FINISHED mechanism seems to work... in this > > case, it allowed he cursor to skip out of the inset, immediately to > > the right of the parentesis, when pressing the END key. But how now do > > I get it to go forward to the next parentesis, i.e., execute LFUN_END > > for the parent inset? (Yes, I know, pressing the END key again. Not > > what I wanted to hear :) > > Maybe we should really have a 'request queue' in the cursor. This would > certainly simplify the logic in this area: > > When dispatching an LFUN_END at pos == lastpos, idx == lastidx, we'd > simply pop off a cursor slice and re-add an LFUN_END request to that > queue. > > LCursor::Dispatch would loop as long as there are events in that quue > and ths pick up this re-added event in the next iteration and > effectively dispatch it to the original inset's parent. No LFUN_FINISHED > needed, no magical cursor.undispatch calls and even more flexible for > the future as an inset may add more than one new event to the queue > if needed etc... > > Andre'
This sounds great... to make it perfectly clear, you propose to replace
the loop at
204 for (; depth(); pop()) {
...
214 disp_.update(true);
215 disp_.dispatched(true);
216 inset().dispatch(*this, cmd);
217 if (disp_.dispatched())
218 break;
219 }
by a similar one testing for
while(!lfun_queue.empty()) {
...
inset().dispatch(*this, cmd);
lfun_queue.pop_back();
}
ands replace the stanzas (pseudo-code, didn't check C++ syntax)
case LFUN_<something>:
<do something>;
cmd = FuncRequest(LFUN_FINISHED_<direction>);
cur.undispatched();
break;
by
case LFUN_<something>:
<do something>;
cur.lfun_queue.push_back(LFUN_<something>));
break;
?
The current solution is indeed a bit of a kludge, as it allows only
queing of a single LFUN to the inset level above.
Perhaps not for 1.4.0 though as we're in freeze.
- Martin
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