SendKeys does not work from pre-defined LPSTR.

Question

I have a QTimer connected to a TimerHandler function. The TimerHandler should do my SendKeys function that I declared in the public section of my header. If I manually type in the text for the SendKeys function it gives an correct output. But if I pass the text from a pre-defined LPSTR it outputs garbage. Here is my code:

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);
}

This makes my timer output garbage instead the text inside txtMessage. If I use

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

instead, it outputs the message correctly. Is there something wrong with my code?

I tried declaring LPSTR txtMessage inside my class in the public section at MyProject.h but this also didn't work.

Answer 1

Make txtMessage a string object (std::string or QString, since you're using Qt), not a pointer. Get the pointer from that object right before making the call to SendText or, even easier, just make SendText take a string object instead of a pointer.

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

The problem is that you're storing a pointer to data in a temporary object (convertme). This object goes out of scope and is destroyed and the memory gets rewritten. The reason it works with `"test message" is that string literals are stored differently. You need to keep the message you're trying to store in memory the whole time.