为什么信号/插槽无法使用多个线程？

Question

class A : public QObject{

  Q_OBJECT

signals:
  void a_sig();

public:
  A(){ }

public slots:
  void begin(){
    QObject::connect(&_timer, SIGNAL(timeout()), this, SIGNAL(a_sig()));
    _timer.start(1000); 
  }

private:
  QTimer _timer;
};


class B : public QObject{

  Q_OBJECT

public:
  B(){ value = 0; }

public slots:
  void b_slot(){

    ++value;
    QFile file("out.txt");
    file.open(QIODevice::WriteOnly);
    QTextStream out(&file);
    out << value << "\n";
    file.close();
  }

private:
  int value;
};

int main(int argc, char **argv){

  QCoreApplication app(argc, argv);

  A a;
  B b;
  QThread aThread;
  QThread bThread;

  QObject::connect(&aThread, SIGNAL(started()), &a, SLOT(begin()));
  QObject::connect(&a, SIGNAL(a_sig()), &b, SLOT(b_slot()));

  a.moveToThread(&aThread);
  b.moveToThread(&bThread);

  aThread.start();
  bThread.start();

  return app.exec();
}

我试图理解为什么b_slot()不被召唤.任何人都可以解释发生了什么,为什么b_slot()不被调用？

Answer 1

问题是班级_timer成员的所有权A.

由于您没有显式初始化它,因此它在没有父对象的情况下进行初始化.因此,a.moveToThread(&aThread)不会将计时器移动到那里aThread,之后就会感到困惑.

将A构造函数更改为:

A() : _timer(this) {}

你的b_slot()意志就会被召集