也许我过于雄心勃勃,但我正在尝试编写一个可以通过QLocalSockets和QTcpSockets接受连接的服务器程序。这个概念是有一个'nexus'对象,QLocalServer和QTcpServer都在监听新的连接:
Nexus::Nexus(QObject *parent)
: QObject(parent)
{
// Establish a QLocalServer to deal with local connection requests:
localServer = new QLocalServer;
connect(localServer, SIGNAL(newConnection()),
this, SLOT(newLocalConnection()));
localServer -> listen("CalculationServer");
// Establish a UDP socket to deal with discovery requests:
udpServer = new QUdpSocket(this);
udpServer -> bind(QHostAddress::Any, SERVER_DISCOVERY_PORT);
connect(udpServer, SIGNAL(readyRead()),
this, SLOT(beDiscovered()));
// Establish a QTcpServer to deal with remote connection requests:
tcpServer = new QTcpServer;
connect(tcpServer, SIGNAL(newConnection()),
this, SLOT(newTcpConnection()));
tcpServer -> listen(QHostAddress::Any, SERVER_COMMAND_PORT);
}
...然后分开建立服务器对象的槽,其构造函数采用指向QIODevice的指针。从理论上讲,这个应该可以工作,因为QLocalSocket和QTcpSocket都继承了QIODevice。这是newLocalConnection插槽,例如:
void Nexus::newLocalConnection()
{
// Create a new CalculationServer connected to the newly-created local socket:
serverList.append(new CalculationServer(localServer -> nextPendingConnection()));
// We don't allow more than one local connection, so stop listening on the server:
localServer -> close();
}
问题是这不会编译,给出错误:
错误C2664: “CalculationServer :: CalculationServer(QIODevice中 *,QObject *)':无法将参数1从'QLocalSocket *'转换为 'QIODevice *'1>类型指向 与...无关;转换需要 reinterpret_cast,C风格演员或 功能式演员
现在指向的类型显然不无关,在我的代码中的其他地方我没有任何问题,例如:
QLocalSocket *socket = new QLocalSocket;
QIODevice *server = new QIODevice;
server = socket;
...所以有人能告诉我为什么编译器有问题吗?有没有办法让构造函数接受QLocalServer *?我想有一个核选项,让构造函数获取一个void指针加上一个额外的变量来告诉它它是什么发送的,所以它可以将void指针重新转换为QLocalSocket或QTcpSocket,但我觉得不舒服求助于reinterpret_cast在看起来它应该是一个简单的C ++多态性。
此致
斯蒂芬。
答案 0 :(得分:1)
最可能的原因是您忘记了发生错误的源文件中的#include <QLocalSocket>
。