On Sat, Dec 12, 2015 at 12:34:47AM +0000, Guillaume Munch wrote:
> Le 11/12/2015 22:47, Enrico Forestieri a écrit :
> >On Fri, Dec 11, 2015 at 06:52:19PM +0000, Guillaume Munch wrote:
[...]
> >>• Defect: They tend to accumulate and they are redundant because there
> >> is only one additional '\n' in the LaTeX source. The problem is the
> >> one can add one before another. Thus, two such consecutive chars
> >> should be merged as a single one, like it is done for spaces in LyX.
> >
> >There were a lot of problems to be solved and this one was escaping me.
> >Thank you for pointing it out. I will try to address it when I will find
> >time (and motivation) to work again on LyX. Unfortunately, I will be
> >unavailable the next weeks.
>
> Thanks.
The attached patch avoids the proliferation of parbreak seaparators.
--
Enrico
diff --git a/src/Text3.cpp b/src/Text3.cpp
index 9edda1b..ca6ec5b 100644
--- a/src/Text3.cpp
+++ b/src/Text3.cpp
@@ -1097,6 +1097,8 @@ void Text::dispatch(Cursor & cur, FuncRequest & cmd)
cap::replaceSelection(cur);
pit_type pit = cur.pit();
Paragraph const & par = pars_[pit];
+ bool lastpar = (pit == pit_type(pars_.size() - 1));
+ Paragraph const & nextpar = lastpar ? par : pars_[pit + 1];
pit_type prev = pit;
if (pit > 0) {
if (!pars_[pit - 1].layout().isEnvironment())
@@ -1105,10 +1107,13 @@ void Text::dispatch(Cursor & cur, FuncRequest & cmd)
prev = pit - 1;
}
if (prev < pit && cur.pos() == par.beginOfBody()
- && !par.isEnvSeparator(cur.pos())
+ && !par.isEnvSeparator(par.size() ? par.size() - 1 : 0)
&& !par.layout().isCommand()
&& pars_[prev].layout() != par.layout()
- && pars_[prev].layout().isEnvironment()) {
+ && pars_[prev].layout().isEnvironment()
+ && (!nextpar.isEnvSeparator(nextpar.size() ? nextpar.size() - 1 : 0)
+ || nextpar.layout().isCommand()
+ || nextpar.layout().isEnvironment())) {
if (par.layout().isEnvironment()
&& pars_[prev].getDepth() == par.getDepth()) {
docstring const layout = par.layout().name();
@@ -1124,6 +1129,8 @@ void Text::dispatch(Cursor & cur, FuncRequest & cmd)
Font const f(inherit_font, cur.current_font.language());
pars_[cur.pit() - 1].resetFonts(f);
} else {
+ if (par.isEnvSeparator(cur.pos()))
+ cur.posForward();
breakParagraph(cur, cmd.argument() == "inverse");
}
cur.resetAnchor();
