我正在将Lambda表达式连接到QObject
的信号:
QObject::connect(handle, &BatchHandle::progressMax, [this](const ProcessHandle* const self, const int value) {
this->maxProgress(value);
});
上面的代码没有任何问题。
然而,Qt::QueuedConnection
是绝对必要的,因为 handle
对象最终会移动到另一个线程。
我将以下内容添加到我的代码中:
QObject::connect(handle, &BatchHandle::finished, [this](const ProcessHandle* const self) {
this->processIsRunning(false);
}, (Qt::ConnectionType)Qt::QueuedConnection);
注意我如何添加显式转换以确保它正确地识别值类型。结果:
1>src\TechAdminServices\database\techCore\processes\import\ImportManagerDialog.cpp(191): error C2664: 'QMetaObject::Connection QObject::connect<void(__cdecl taservices::ProcessHandle::* )(const taservices::ProcessHandle *),Qt::ConnectionType>(const taservices::ProcessHandle *,Func1,const QObject *,Func2,Qt::ConnectionType)' : cannot convert parameter 3 from 'taservices::`anonymous-namespace'::<lambda58>' to 'const QObject *'
1> with
1> [
1> Func1=void (__cdecl taservices::ProcessHandle::* )(const taservices::ProcessHandle *),
1> Func2=Qt::ConnectionType
1> ]
1> No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called
如何在连接Lambda时获取排队的连接?