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Qt signals and slots, threads, app.exec(), and related queries

[related to this question]

I wrote this piece of code to understand how qt signals and slots work. I need someone to explain the behaviour, and to tell me if I'm right about my own conclusions.

My program:

connectionhandler.h

#ifndef CONNECTIONHANDLER_H
#define CONNECTIONHANDLER_H

#include <QTcpServer>
class ConnectionHandler : public QObject
{
    Q_OBJECT
public:
    ConnectionHandler();
public slots:
    void newConn();
private:
    QTcpServer *server;
};

#endif // CONNECTIONHANDLER_H

connectionhandler.cpp

#include "connectionhandler.h"
#include <QTextStream>

ConnectionHandler::ConnectionHandler() {
    server = new QTcpServer;
    server->listen(QHostAddress::LocalHost, 8080);
    QObject::connect(server, SIGNAL(newConnection()),this, SLOT(newConn()));
}
void ConnectionHandler::newConn() {
    QTextStream out(stdout);
    out << "new kanneksan!\n";
    out.flush();
}

main.cpp

#include <QCoreApplication>
#include "connectionhandler.h"

int main(int argc, char* argv[]) {
    QCoreApplication app(argc,argv);
    ConnectionHandler handler;
    return app.exec();
}

Now, running this program sends it into an infinite loop looking for new connections.

Observation: if I don't call app.exec(), the program returns immediately (as it should).
Question: why?

Question: if I had connected the slot as a queued connection, when would the slot invocation be performed?
Question: if app.exec() is an infinite loop of sorts, how does the newConnection() signal ever get emitted?

Big Question: Is their any "second thread" involved here? (I expect a no, and a stunningly elegant explanation :) )

Thanks,
jrh

PS: who else has this nested parenthesis syndrome? like "(.. :))" or "(.. (..))"?

like image 589
jrharshath Avatar asked Sep 24 '09 14:09

jrharshath


1 Answers

If you don't call app.exec() then the program hits the end of your main() and ends. (Why? There's no more code to execute!)

app.exec() is an infinite loop of the following style:

do
{
  get event from system
  handle event
}
while (true);

If you use a queued connection, then the event is added to your event queue, and it will be performed at some point in the future during the app.exec() loop.

There is no second thread in your program. Events are delivered asynchronously by the OS, which is why it appears that there's something else going on. There is, but not in your program.

like image 197
Bill Avatar answered Sep 28 '22 00:09

Bill