Better return nil than an arbitrary xml:lang="..." value

...for a localized property that doesn't have a suitable value, but is nillable.

E.g., /org.openoffice.Office.Writer/Insert/Caption/CaptionOrderNumberingFirst
(once fixed to indeed be a localized property as apparently intended, with a
follow-up commit to this) only has an explicit value (true) for xml:lang="hu",
but shouldn't return that for any other locale (where the implicit default for a
nil value should instead be false, cf. the else branch at the end of
SwInsertConfig::Load, sw/source/uibase/config/modcfg.cxx).

Change-Id: I7b991c1bc4df22bf2359175b0734e85e0844ce99
Reviewed-on: https://gerrit.libreoffice.org/49409
Tested-by: Jenkins <ci@libreoffice.org>
Reviewed-by: Stephan Bergmann <sbergman@redhat.com>

1 files changed, 7 insertions, 6 deletions

diff --git a/configmgr/source/access.cxx b/configmgr/source/access.cxx
index 0d19af353b49..718f6339967b 100644
--- a/configmgr/source/access.cxx
+++ b/configmgr/source/access.cxx
@@ -1430,9 +1430,8 @@ rtl::Reference< ChildAccess > Access::getChild(OUString const & name) {
                 }
             }
         }
-        // Defaults are the "en-US" locale, the "en" locale, the empty string
-        // locale, the first child (if any), or a null ChildAccess, in that
-        // order:
+        // Defaults are the "en-US" locale, the "en" locale, the empty string locale, the first child (if
+        // any, and if the property is non-nillable), or a null ChildAccess, in that order:
         rtl::Reference< ChildAccess > child(getChild("en-US"));
         if (child.is()) {
             return child;
@@ -1445,9 +1444,11 @@ rtl::Reference< ChildAccess > Access::getChild(OUString const & name) {
         if (child.is()) {
             return child;
         }
-        std::vector< rtl::Reference< ChildAccess > > children(getAllChildren());
-        if (!children.empty()) {
-            return children.front();
+        if (!static_cast<LocalizedPropertyNode *>(getNode().get())->isNillable()) {
+            std::vector< rtl::Reference< ChildAccess > > children(getAllChildren());
+            if (!children.empty()) {
+                return children.front();
+            }
         }
         return rtl::Reference< ChildAccess >();
     }