Question

我有QTimer与TimerHandler功能相关联。 TimerHandler应该执行我在标题的公共部分中声明的SendKeys函数。如果我手动输入SendKeys函数的文本，它会给出正确的输出。但是，如果我从预定义的LPSTR传递文本，它会输出垃圾。这是我的代码：

MyProject.h

#ifndef MYPROJECT_H
#define MYPROJECT_H

#include <QtGui/QMainWindow>
#include "ui_myproject.h"
#include <qtimer.h>
#include <qmessagebox.h>
#include <Windows.h>

class MyProject : public QMainWindow
{
    Q_OBJECT

public:
    MyClass(QWidget *parent = 0, Qt::WFlags flags = 0);
    Ui::MyProjectClass ui;
    QTimer* SpamTimer;

    void SendText(char* message, int size)
    {
        int lc=0;
        do{
        keybd_event(VkKeyScan(message[lc]),0,KEYEVENTF_EXTENDEDKEY,0);
        keybd_event(VkKeyScan(message[lc]),0,KEYEVENTF_KEYUP,0);
        lc=lc+1;
        }while(lc<size);
        keybd_event(VK_RETURN,0,KEYEVENTF_EXTENDEDKEY,0);
        keybd_event(VK_RETURN,0,KEYEVENTF_KEYUP,0);
    }

public slots:
    void StartBTNClick();
    void StopBTNClick();
    void TimerHandler();
};
#endif // MYPROJECT_H

MyProject.cpp

#include "MyProject.h"

LPSTR txtMessage; // Message for SendKeys function.
int buffer;
bool TimerEnabled = 0;

MyClass::MainWindow(QWidget *parent, Qt::WFlags flags) // Intializing MainWindow
    : QMainWindow(parent, flags)
{
    ui.setupUi(this);
    statusBar()->showMessage("Status: Idle.");
    connect(ui.StartBTN, SIGNAL(clicked()), this, SLOT(StartBTNClick()));
    connect(ui.StopBTN, SIGNAL(clicked()), this, SLOT(StopBTNClick()));
}

void MyClass::StartBTNClick() // Starts the timer.
{
    int delay; // delay for QTimer
    bool ok;
    std::string convertme;

    QString TextMSG = ui.TextBox->text(); // Get text from 'line edit' for txtMessage.
    QString TimeMSG = ui.TimeBox->text(); // Get text from 2nd 'line edit' for delay.
    convertme = TextMSG.toStdString();
    txtMessage = const_cast<char*> (convertme.c_str()); // converted QString to LPSTR.
    buffer = strlen(txtMessage);
    delay = TimeMSG.toInt(&ok, 10); // converted QString to int.
    if (delay > 0)
    {
        QtTimer = new QTimer(this);
        connect(QtTimer, SIGNAL(timeout()), this, SLOT(TimerHandler()));
        TimerEnabled = 1;
        QtTimer->start(delay);
        statusBar()->showMessage("Status: Running.");
    }
    else if (delay < 0)
    {
        QMessageBox::warning(this, "Warning!", "Delay can't be \"0\" or lower than \"0\"!");
    }
    else
    {
        QMessageBox::warning(this, "Warning!", "Delay was not specified properly.");
    }
}

void MyClass::StopBTNClick() // Stops the timer.
{
    if (TimerEnabled == 1)
    {
        QtTimer->stop();
        disconnect(QtTimer, SIGNAL(timeout()), this, SLOT(TimerHandler()));
        TimerEnabled = 0;
        statusBar()->showMessage("Status: Idle.");
    }
}

void MyClass::TimerHandler() // Timer handles the SendKeys function
{
    SendText(txtMessage, buffer);
}

这使我的计时器输出变为垃圾而不是txtMessage内的文本。如果我使用

SendText("test message", strlen("test message"));

相反，它正确输出消息。我的代码有问题吗？

我尝试在LPSTR txtMessage的公开栏目中在我的班级内宣布MyProject.h，但这也无效。

Answer 1

让txtMessage成为string对象（std::string或QString，因为您使用的是Qt），而不是指针。在调用SendText之前获取该对象的指针，或者更简单，只需使SendText取一个字符串对象而不是指针。

void SendText(const QString& str)
{
  const char* message = str.c_str();
  // whatever else you want to do
}

问题是您在临时对象（convertme）中存储指向数据的指针。此对象超出范围并被销毁，内存被重写。它与“测试消息”一起使用的原因是string literals are stored differently。您需要始终保留您尝试存储在内存中的消息。

SendKeys不能使用预定义的LPSTR。

1 个答案: