Error while using QTcpSocket - c++

Question

I am creating a (very basic) MJPG server to show webcam data on a browser. I have partly managed to do it so far.
Here is my code:
TcpServer::TcpServer(QObject *parent) :
QObject(parent)
{
server = new QTcpServer(this);
// whenever a user connects, it will emit signal
connect(server, SIGNAL(newConnection()),
this, SLOT(newConnection()));
if (!server->listen(QHostAddress::Any, 9999))
qDebug() << "Server could not start";
else
qDebug() << "Server started!";
}
...
void TcpServer::newConnection()
{
QTcpSocket *socket = server->nextPendingConnection();
QByteArray ContentType = ("HTTP/1.0 200 OK\r\n" \
"Cache-Control: no-cache\r\n" \
"Cache-Control: private\r\n" \
"Content-Type: multipart/x-mixed-replace;boundary=--boundary\r\n");
socket->write(ContentType);
std::vector<uchar> buff;
Mat img; //OpenCV Material
while (1) {
// Image to Byte Array via OPENCV Method
buff.clear();buff.empty();
vCapture->read(img); //Read from webcam
imencode(".jpg", img, buff, compression_params);
std::string content(buff.begin(), buff.end());
QByteArray CurrentImg(QByteArray::fromStdString(content));
QByteArray BoundaryString = ("--boundary\r\n" \
"Content-Type: image/jpeg\r\n" \
"Content-Length: ");
BoundaryString.append(QString::number(CurrentImg.length()));
BoundaryString.append("\r\n\r\n");
socket->write(BoundaryString);
socket->write(CurrentImg); // Write The Encoded Image
socket->flush();
}
}
The Problem -
When I run this program, the first image is shown. After that, the following error is continuously printed on the app -
QIODevice::write (QTcpSocket): device not open
It looked like the socket got closed, so I used re-initialized the socket, like this - socket = server->nextPendingConnection();, although the app threw an error with this code. Any help on how to fix this?
EDIT -
I tried the lambda method and it worked fine. However, I still face 2 problems -
The image size has to be excessively low (around 270x480 with lowest JPG quality)
(MORE IMPORTANT) I have to manually press the reload button on browser to reload the image, it doesn't automatically change from one image to the other.

Answer 1

It looked like the socket got closed
Rather than guessing, connect to the error signals of TCPServer and TCPSocket to know when errors occur, or when a client disconnects.
The problem you have is the while(1) loop. Qt is an event-driven framework, so having code in an infinite loop on the main thread is going to prevent events being delivered.
Instead of the infinite loop, connect to QTcpSocket::readyRead signal and handle the data when the connected slot is called.
Qt demonstrates this with the Fortune Server and Fortune Client example code.
If you're using C++ 11, you can use a connection to a lambda function to handle the readyRead, like this
void TcpServer::newConnection()
{
...
QTcpSocket *m_TcpHttpClient = server->nextPendingConnection();
connect(m_TcpHttpClient, &QTcpSocket::readyRead, [=](){
// handle received data here
});
}

Answer 2

Here Is the Code I used to open the MJPEG Streaming Server in my original Project. Perhaps It can help you finding out your Problem.
void MjpegStreamingEngine::StartServer(){
m_TcpHttpServer = new QTcpServer();
m_TcpHttpServer->connect(m_TcpHttpServer,
SIGNAL(newConnection()),
this,
SLOT(TcpHttpconnected()));
m_TcpHttpServer->listen(QHostAddress::Any,
8889);
}
void MjpegStreamingEngine::TcpHttpconnected(){
m_TcpHttpClient = m_TcpHttpServer->nextPendingConnection();
m_TcpHttpClient->connect(m_TcpHttpClient,
SIGNAL(readyRead()),
this,
SLOT(TcpHttpreadyRead()));
}
void MjpegStreamingEngine::TcpHttpreadyRead(){
m_TcpHttpClient->readAll(); // Discard "Get Request String"
QByteArray ContentType = ("HTTP/1.0 200 OK\r\n" \
"Server: en.code-bude.net example server\r\n" \
"Cache-Control: no-cache\r\n" \
"Cache-Control: private\r\n" \
"Content-Type: multipart/x-mixed-replace;boundary=--boundary\r\n\r\n");
m_TcpHttpClient->write(ContentType);
while(1){
if (m_TcpHttpClient->isOpen())
{
// Send Image
QThread::msleep(10);
}else{
return;
}
}
m_TcpHttpClient->disconnect(); //Should never be Reached
}