[partial solved] How emit SIGNAL

Franckynos

Hi,

I have a problem with my code.
I use some card "Yoctopuce" write in c++.
The library work with callback.
I want emit signal when callback arrived.
For define a callback i do this :

YVoltage*  volt = YVoltage::FindVoltage(target + ".voltage");
volt->registerValueCallback(mycall);

void mycall(YSensor *volt, const string& value)
{
    qDebug() << "plop" << volt->get_module()->get_serialNumber().c_str() <<value.c_str();
    //emit COCO();
}

but my function call (callback) is not member of my Qt class so I can't emit signals... who can I do this ?

The parameters of registerValueCallback is (virtual int registerValueCallback(YVoltageValueCallback callback);)
and YVoltageValueCallback is (typedef void (*YVoltageValueCallback)(YVoltage *func, const string& functionValue);)

Someone can help me ?

jsulm

If I understand you right you already have a class derived from QObject. In this case you can use an instance of that class to emit a signal:

void mycall(YSensor *volt, const string& value)
{
    myClassInstance.foundVoltage();
}

void MyClass::foundVoltage()
{
   emit mySignal();
}

Franckynos

@jsulm said:

myClassInstance

not really I have no access to QObject or Q ... something in mycall, it's a "volatile" function.

it's a callback so pointer of function.

If I try to insert mycall in My class the library yoctopuce says:

Impossible to convert Motor::* in YVoltageValueCallback

volt->registerValueCallback(&Motor::mycall)

jsulm

@Franckynos But in your function you can call any public methods of any class instance, you just need access to such an instance.

Hamed.Masafi

You can implement observer pattern.

Franckynos

@jsulm

but the function is static but I'll tried to do something with pointer...

Franckynos

@jsulm I have declared a pointer to my class to have access at the instance and it works! I found this methode not really good with static member but for the moment I have juste this solution.