Question

我想从“ takeSnapshots”类中创建对象，该对象在实例化后将学习“ lock”类中另一个对象的名称，以便当“ lock”对象的状态改变时可以稍后查询。我可以想到多种让类“ takeSnapshots”中的对象知道要查询的对象的方式（例如，将“ lock”对象的名称作为对其成员函数的调用的一部分）。但是，我认为最好一开始就处理好关系，以后再担心是否调用了正确的对象组合。

所包含的代码显示了两个类的简化版本以及在main中创建的示例实例。我已经在它们各自的提示之后的每一行中包含了输出。我所期望的是，“ takeSnapshots”的构造函数将存储“ lock”对象的地址。然后我可以在以后拍摄快照时使用它。您可以看到，存储的内容（至少当我使用它获取numWheels时）与“锁定”对象认为它具有numWheels的地址有一些距离。

我最想知道为什么这段代码不能按我期望的方式工作，也就是说，如果这不是一个好的架构思想，那是一回事。但是鉴于我在这里看到的行为，我显然不准备在任何复杂的情况下使用指针，并且我不想因为错误的实现而放弃基本架构。任何帮助将不胜感激。

// simpleTest.cpp : Demonstrates problem I'm having understanding 
// pointer to an object.

#include "stdafx.h"
#include<iostream>
using namespace std;

class lock {
public:     //Just while I run a test.
    int numWheels;
    lock(int numW) {
        numWheels = numW;
        cout << " \n In \"lock\" constuctor, address and value of numWheels 
          " << &numWheels << " " << numWheels << endl;
    }   //Values from console: 0034F874 and 4
};


class takeSnapshots {
    lock* localLock;
    public:
    takeSnapshots(lock myLock) {
        localLock = &myLock;
        cout << " \n In \"takeSnapshots\" constuctor, address and value of 
             numWheels " << &localLock->numWheels << " " 
             << localLock->numWheels << endl;
    //Values from console: 0034F794 and 4  "Same value, but not the same 
     //address as expected from "lock."
    }
    void takeASnapSnapshot() {
        cout << " \n When taking a snapshot, address and value of numWheels 
          " << &localLock->numWheels << " " << localLock->numWheels << 
            endl;
    //Values from console: 0034F794 and 2303449     "No longer even the 
    // same value as expected from "lock."
    }
};

int main()
{
    lock yourLock(4);
    takeSnapshots myShots1(yourLock);

    cout << " \n In main (from \"yourLock\"), address and value of 
         numWheels " << &yourLock.numWheels << " " << yourLock.numWheels << 
         endl;
//Values from console: 0034F874 and 4       "Still the same values as set 
//in the constructor of "lock."

//Take a picture
        myShots1.takeASnapSnapshot();

    return 0;
}

在构造另一个对象期间存储对象名称

0 个答案: