QWinEventNotifier：只能用于用QThread启动的线程

Question

我想用“QextSerialPort”打开我的Huwawei USB加密狗。

我的端口详情如下

Port Name: 
Product ID: 
Physical Name: \Device\000000ca 
Vendor Id: 
Friend Name: SB 

Port Name: 
Product ID:? 
Physical Name: \Device\USBPDO-10 
Vendor Id: ? 
Friend Name: TH 

Port Name: COM3 
Product ID: 
Physical Name: \Device\BthModem0 
Vendor Id: 
Friend Name: Standard Serial over Bluetooth link (COM3) 

Port Name: COM4 
Product ID: 
Physical Name: \Device\BthModem2 
Vendor Id: 
Friend Name: Standard Serial over Bluetooth link (COM4) 

Port Name: COM5 
Product ID: 
Physical Name: \Device\BthModem1 
Vendor Id: 
Friend Name: Standard Modem over Bluetooth link 

Port Name: COM6 
Product ID:? 
Physical Name: \Device\000000e2 
Vendor Id: ? 
Friend Name: HUAWEI Mobile Connect - 3G Application Interface (COM6) 

Port Name: COM7 
Product ID:? 
Physical Name: \Device\000000e0 
Vendor Id: ? 
Friend Name: HUAWEI Mobile Connect - 3G Modem 

Port Name: COM8 
Product ID:? 
Physical Name: \Device\000000e3 
Vendor Id: ? 
Friend Name: HUAWEI Mobile Connect - 3G PC UI Interface (COM8) 

我试图打开我的USB加密狗，这样我就可以发送短信。以下是我开

#include "MyClass.h" 
#include <qstring.h> 
#include <qdebug.h> 


int main() 
{ 
    QextSerialPort *port = new QextSerialPort("COM7"); 
    port->open(QIODevice::ReadWrite); 
    cout << port->isOpen(); 

    system("pause"); 
    return 0; 

} 

代码，当我运行这段代码，我得到的是

QWinEventNotifier: Can only be used with threads started with QThread 
1 

这显示了端口ID打开，但对于该消息？这是否意味着我无法继续使用其他代码？在我编写任何其他代码之前，我想知道这一点。请帮忙！

Answer 1

最有可能的，你需要创建一个QApplication，没有它一样的事件和信号许多事情/插槽将无法正常工作：

int main() 
{ 
    QApplication app; 

    QextSerialPort *port = new QextSerialPort("COM7"); 
    port->open(QIODevice::ReadWrite); 
    cout << port->isOpen(); 

    system("pause"); 

    return app.exec(); 
}