Question

我考虑使用QProcess多次调用命令行应用程序（http://plnkr.co/edit/RxTte5njdS8GfTiD73LA?p=preview）。每次用户单击按钮，都会发出命令。

将监控应用输出并重定向到屏幕。代码如下所示。

void Gpio::command(QString argument)
{
//    if(process)
//        delete process;
    process = new QProcess(this);
    connect(process, SIGNAL(started()), this, SLOT(onStart()));
    connect(process, SIGNAL(finished(int,QProcess::ExitStatus)), this, SLOT(onFinish(int,QProcess::ExitStatus)));
    connect(process, SIGNAL(readyReadStandardOutput()), this, SLOT(readGpio()));
    QString program("gpio");
    QStringList list = argument.split(" ");
    process->start(program, list);
}

问题：我应delete process吗？这样做我得到了：

QProcess: Destroyed while process is still running.

监控exitCode和exitStatus我发现它们总是0。

此问题更多地涉及正确使用QProcess，gpio关注特定错误。

Answer 1

由于您不希望同时运行多个进程（根据注释），因此您无需多次创建/删除QProcess。

gpio.h

QProcess*   m_gpioProcess;

gpio.cpp文件

Gpio::Gpio(.....),
   .....(),
   m_gpioProcess(new QProcess(this))
{
   m_gpioProcess->setProgram("gpio");

   connect(m_gpioProcess, SIGNAL(started()), this, SLOT(onStart()));
   connect(m_gpioProcess, SIGNAL(finished(int,QProcess::ExitStatus)), this, SLOT(onFinish(int,QProcess::ExitStatus)));
   connect(m_gpioProcess, SIGNAL(readyReadStandardOutput()), this, SLOT(readGpio()));
}

void Gpio::command(const QString& args)
{
  if (m_gpioProcess->state() != QProcess::NotRunning) {
    qDebug() << "Process already running, ignoring the request";
    return;
  }

  m_gpioProcess->setArguments(args.split(" "));
  m_gpioProcess->start();
  if (m_gpioProcess->waitForStarted()) {
    qDebug() << "Process started with arguments:" << m_gpioProcess->arguments();
  }
}

如果您想阻止用户多次点击该按钮，请考虑按m_gpioProcess状态启用/禁用该按钮。

对于Qt 4.8，只需删除此行

即可

m_gpioProcess->setProgram("gpio");

和这一行

m_gpioProcess->setArguments(args.split(" "));

并更改此行

m_gpioProcess->start();

到

m_gpioProcess->start("gpio", args.split(" "));

Answer 2

好吧，在调用process->waitForFinished();之后，我似乎忘记了start。这似乎解决了这个问题。

Answer 3

添加以下代码可能会有帮助：

process->close();

正确使用Qt QProcess

3 个答案: