isSignalConnected correct usage

pixbyte

to check if a signal is connected I use this code:

    if (!this->isSignalConnected(QMetaMethod::fromSignal(&QtVariantPropertyManager::propertyChanged))) {

        connect( sdkItemPropManager, SIGNAL( propertyChanged( QtProperty * ) ), this, SLOT( onUpdatePropertyValues() ) );
        connect( sdkEmuManager, SIGNAL( propertyChanged( QtProperty * ) ), this, SLOT( onUpdatePropertyValues() ) );

    }

But I always got an Assert:

ASSERT failure in QObject::isSignalConnected: "the parameter must be a signal member of the object", file \Users\qt\work\qt\qtbase\src\corelib\kernel\qobject.cpp, line 2539

Is there an explanation that you can understand?

Christian Ehrlicher

Why not simply use Qt::UniqueConnection?

According to the documentation: "signal must be a signal member of this object, otherwise the behaviour is undefined."

pixbyte

@Christian-Ehrlicher said in isSignalConnected correct usage:

Qt::UniqueConnection

ok, this is really a good and helpfull tip. Thank you. But also Iam interrested why the initial code do not work.

JonB

@pixbyte
I assume as the error message states.

this->isSignalConnected(QMetaMethod::fromSignal(&QtVariantPropertyManager::propertyChanged)

Is this of type QtVariantPropertyManager?

pixbyte

No this is the target where the signal is connected on. I think here it is a mdi_child.

JonB

@pixbyte
Hence I presume the message

"signal must be a signal member of this object, otherwise the behaviour is undefined."

in answer to your

But also Iam interrested why the initial code do not work.