当Activity启动时，EditText具有焦点，但不显示软键盘(EditText has focus when Activity starts, but soft keyboard is not shown)

 如果showScreen中声明为Qt Slot ，例如： 
 
private slots:
    void showScreen(QWidget* w);
 
 并且您的信号在buttons_widget中声明 
 
signals:
    void showPositionScreen_signal(QWidget* w); //Note that signal needs same type as slot
    void showTimingScreen_signal(QWidget* w);
 
 然后，您可以将该信号连接到插槽。 请注意，信号和插槽的参数必须匹配。 即： “信号和时隙机制是类型安全的：信号的签名必须与接收时隙的签名匹配。”实际上，一个时隙可能比它收到的信号具有更短的签名，因为它可以忽略额外的参数。“ 
 
connect(buttons_widget, SIGNAL(showPositionScreen_signal(QWidget*)), this, SLOT(showScreen(QWidget*)));
 
 你必须从timing_screen发出position_screen和timing_screen ，如： 
 
emit showPositionScreen_signal(position_screen);
 
 正如thuga所指出的，这就是说你不需要两种不同的信号。 要将另一个QWidget传递到同一个插槽，只需用它发出该信号即可。 即： 
 
emit showPositionScreen_signal(timing_screen);
 
 我建议将信号的名称改为适当的名称。 

If showScreen is declared as Qt Slot in your mainwindow.h like:
 
private slots:
    void showScreen(QWidget* w);
 
And your Signals are declared in buttons_widget
 
signals:
    void showPositionScreen_signal(QWidget* w); //Note that signal needs same type as slot
    void showTimingScreen_signal(QWidget* w);
 
Then you can connect that signal to a slot. Note that the arguments of signals and slots have to match. I.e: "The signals and slots mechanism is type safe: The signature of a signal must match the signature of the receiving slot. (In fact a slot may have a shorter signature than the signal it receives because it can ignore extra arguments.)"
 
connect(buttons_widget, SIGNAL(showPositionScreen_signal(QWidget*)), this, SLOT(showScreen(QWidget*)));
 
And you will have to emit position_screen and timing_screen from within buttons_widget like:
 
emit showPositionScreen_signal(position_screen);
 
As thuga pointed out, it is to say that you do NOT need two different signals. To pass another QWidget to the same slot simply emit that signal with it. I.e.:
 
emit showPositionScreen_signal(timing_screen);
 
And i would suggest to change the name of your signal to something appropriate then.