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authorJ-P Nurmi <jpnurmi@theqtcompany.com>2015-09-14 21:23:32 +0200
committerJ-P Nurmi <jpnurmi@theqtcompany.com>2015-09-21 19:59:27 +0000
commitb7738beda651c2927e1a9d58c592148b1dc99576 (patch)
tree55d62516410088f22885ff551d57fff5ae50a5bc /src/qml/qml/qqmltypewrapper.cpp
parentba7edffda3f360955825c3ef886ec232c0b2022b (diff)
Make QML composite types inherit enums
Problem: in Qt Quick Controls 2, enums declared in the abstract C++ base types were not accessible with the concrete QML type name, but had to be referenced using the base type name: Slider { snapMode: AbstractSlider.SnapOnRelease } Solution: this change resolves the C++ base type and creates the missing link between the composite type and its base type's meta- object. This allows referencing enums using the concrete/composite QML type name: Slider { snapMode: Slider.SnapOnRelease } Change-Id: Icefdec91b012b12728367fd54b4d16796233ee12 Task-number: QTBUG-43582 Reviewed-by: Simon Hausmann <simon.hausmann@theqtcompany.com>
Diffstat (limited to 'src/qml/qml/qqmltypewrapper.cpp')
-rw-r--r--src/qml/qml/qqmltypewrapper.cpp2
1 files changed, 1 insertions, 1 deletions
diff --git a/src/qml/qml/qqmltypewrapper.cpp b/src/qml/qml/qqmltypewrapper.cpp
index 1d72b2da0d..1145f1b64f 100644
--- a/src/qml/qml/qqmltypewrapper.cpp
+++ b/src/qml/qml/qqmltypewrapper.cpp
@@ -182,7 +182,7 @@ ReturnedValue QmlTypeWrapper::get(const Managed *m, String *name, bool *hasPrope
if (name->startsWithUpper()) {
bool ok = false;
- int value = type->enumValue(name, &ok);
+ int value = type->enumValue(QQmlEnginePrivate::get(v4->qmlEngine()), name, &ok);
if (ok)
return QV4::Primitive::fromInt32(value).asReturnedValue();