C++ Question
What exactly happens with QTimer start function?
I have the following code:
mytimer.cpp
#include "mytimer.h"
#include <QtCore>
MyTimer::MyTimer()
{
timer = new QTimer(this);
connect(timer,SIGNAL(timeout()),this,SLOT(mySlot()));
timer->start(1000);
}
void MyTimer::mySlot()
{
qDebug()<<"timer executed";
}
and in the main.cpp
#include <QCoreApplication>
#include "mytimer.h"
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
MyTimer mtimer;
qDebug()<<"DONE";
return a.exec();
}
Now the output comes as:
DONE
timer executed
timer executed
...
...
...
...
infinite sequence
I am really confused by this. How is that we are done with our main function and still the code for the
SLOT mySlot()What are the important aspects to this? that I need to understand?
Also what changes occurs when I modify mytimer.cpp MyTimer() to:
MyTimer::MyTimer()
{
timer = new QTimer(this);
QEventLoop eventloop;
connect(timer,SIGNAL(timeout()),this,SLOT(mySlot()));
connect(timer,SIGNAL(timeout()),&eventloop,SLOT(quit()));
timer->start(1000);
eventloop.exec();
}
There is one timer executed just before DONE is printed. To Be specific the output now becomes:
timer executed
DONE
timer executed
timer executed
...
...
...
...
infinite sequence
What caused that separate timer executed to appear above the DONE?
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Answer Source
No - your main function is not done. It called a.exec() which will never return in your application.
a.exec() in turn processes a "message queue" which triggers all the timer events that call mySlot().
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