Question

喜欢标题。

我的目标是在不重新创建对象的情况下获取对象，显然我可以使用指针，但

1）我喜欢保留工厂对象，数据库处理程序等的指针

2）我想学习一些东西而不是简单的方法

所以代码和错误都在底部，感谢您的时间和建议：）

代码中的某处

QTcpSocket* socket = (QTcpSocket*)this->sender();
Request request(socket);
Response response(request, this->managerFactory);

response.h

#ifndef RESPONSE_H
#define RESPONSE_H

#include <QString>
#include "Request.h"
#include "../manager/ManagerFactory.h"

class Response {
public:
    Response(const Request &request, ManagerFactory *managerFactory);
    QString getValue();
private:
    QString generateJsonHeader();
    Request &request;
    ManagerFactory *managerFactory;
};

#endif  /* RESPONSE_H */

resonse.cpp

#include "Response.h"

Response::Response(const Request &request, ManagerFactory *managerFactory)
{
    this->request = request;
    this->managerFactory = managerFactory;
}

request.h

#ifndef REQUEST_H
#define REQUEST_H

#include <QTcpSocket>
#include <QStringList>
#include <QString>
#include <QMap>

class Request {
public:
    Request(QTcpSocket* socket);
    bool hasParameter(QString paramName);
    QString getParameter(QString paramName);

private:
    void loadParameters(QTcpSocket* socket);
    QMap<QString, QString> parameters;
};

#endif  /* REQUEST_H */

request.cpp

#include "Request.h"

Request::Request(QTcpSocket* socket)
{
    this->loadParameters(socket);
}

这是我面临的错误

source/server/Response.cpp: In constructor ‘Response::Response(const Request&, ManagerFactory*)’:
source/server/Response.cpp:3:1: error: uninitialized reference member ‘Response::request’ [-fpermissive]
make[2]: *** [build/Release/GNU-Linux-x86/Response.o] Error 1

Answer 1

您正尝试对引用成员变量进行赋值，而不是在初始化列表中初始化它。这是非法的，这就是你遇到错误的原因。

您应该在标题中编写以下构造函数：

class Request;

// ...

class Response {
public:
    Response(Request &r) : request(r) {}

    // ...

private:
    Request &request;
};

虽然，您可能希望考虑使用指针而不是引用成员变量。

无法传递对象作为参考

1 个答案: