C++ 指针解引用问题

我有一个函数

void MOVE\u TO（Square**sendingSquare，Square**receivingSquare）

，其中

Square

是一个类：

class Square;
class Entity
{    
public:
    Square *currentSq;
};//  Just a data type--pretend it's your favorite class.
class Square
{
public:
    Entity *occupant;
};

void MOVE_TO(Square **sendingSquare, Square **receivingSquare)
{
    Entity *movingOccupant = (*sendingSquare)->occupant;
    (*receivingSquare)->occupant = movingOccupant;
    movingOccupant->currentSq = *receivingSquare;
    (*sendingSquare)->occupant = NULL;
}

---

问题是，当

MOVE_TO（…）

返回时，两个方块都指向接收方块，原本应该被移动的居住者完全消失了。想法/建议？我的代码被卡住了，直到我能解决这个问题。如果我在收到任何建议之前就解决了问题，我会回来回答我自己的问题。

我假设您希望将实体的实例从sendingSquare移动到receivingSquare，并调整移动实体的CurrentSquar，使其与receivingSquare一致

如果是这种情况，下面的代码将执行此操作：

#include <iostream>
using namespace std;

class Square;
class Entity
{    
public:
    Square *currentSq;
    Entity(): currentSq(NULL){}//set currentSq to NULL using constructor initialization list
};

class Square
{
public:
    Entity *occupant;
    Square(): occupant(NULL){}//set occupant to NULL using constructor initialization list
};

//MOVE_TO with single '*'
void MOVE_TO(Square *sendingSquare, Square *receivingSquare)
{
    Entity *movingOccupant = sendingSquare->occupant;
    receivingSquare->occupant = movingOccupant;
    movingOccupant->currentSq = receivingSquare;
    sendingSquare->occupant = NULL;
}

int main(int argc, char** argv) {
    //create instances
    Square *sendingSquare = new Square(), *receivingSquare = new Square();
    Entity *entity = new Entity();

    //set up instances accordingly
    sendingSquare->occupant = entity;
    entity->currentSq = sendingSquare;

    //print instances address before MOVE_TO invoked
    //we know that receivingSquare.occupant is NULL, printing receivingSquare.occpuant.currentSq is commented
    cout << "sendingSquare: "<< sendingSquare 
         << ", sendingSquare.occupant: " << sendingSquare->occupant
         << ", sendingSquare.occupant.currentSq: " <<  sendingSquare->occupant->currentSq
         << ", receivingSquare: " <<receivingSquare 
         << ", receivingSquare.occupant: " << receivingSquare->occupant 
         //<< ", sendingSquare.occupant.currentSq: " << receivingSquare.occupant->currentSq
         << endl;

    MOVE_TO(sendingSquare,receivingSquare);

    //print instances address afer MOVE_TO invoked
    //we know that sendingSquare.occupant is NULL, printing sendingSquare.occpuant.currentSq is commented 
    cout << "sendingSquare: "<< sendingSquare 
         << ", sendingSquare.occupant: " << sendingSquare->occupant
         //<< ", sendingSquare.occupant.currentSq: " <<  sendingSquare.occupant->currentSq
         << ", receivingSquare: " << receivingSquare 
         << ", receivingSquare.occupant: " << receivingSquare->occupant 
         << ", receivingSquare.occupant.currentSq: " << receivingSquare->occupant->currentSq
         << endl;

    //commenting instance deletion. The program is ended anyway
    //delete entity,sendingSquare,receivingSquare;
    return 0;
}

#包括
使用名称空间std；
班级广场；
类实体
{    
公众：
平方*当前平方；
Entity（）：currentSq（NULL）{}//使用构造函数初始化列表将currentSq设置为NULL
};
阶级广场
{
公众：
实体*占用人；
Square（）：占用者（NULL）{}//使用构造函数初始化列表将占用者设置为NULL
};
//使用单个“*”将_移动到
无效移动到（正方形*发送正方形，正方形*接收正方形）
{
实体*movingOccupant=sendingSquare->occupant；
接收广场->占用者=移动占用者；
movingOccupant->currentSq=接收广场；
发送方->占用方=空；
}
int main（int argc，字符**argv）{
//创建实例
Square*sendingSquare=new Square（），*receivingSquare=new Square（）；
实体*实体=新实体（）；
//相应地设置实例
发送方->占用方=实体；
实体->当前SQ=发送方；
//在将_移动到之前打印实例地址
//我们知道receivingSquare.occpuant.currentSq为空，printing receivingSquare.occpuant.currentSq被注释
你为什么使用
Square**
指针而不是简单的
Square*
指针作为
MOVE\u TO
参数，这是完全不清楚的，但是如果做得好，它仍然可以工作。它应该以当前的形式工作。我在
MOVE\u TO
中没有看到任何东西会产生你描述的奇怪行为。总之，调用
MOVE_to
？如何调用？我会将
MOVE_to
上的大小写更改为
。它看起来像一个宏，因为宏通常都是大写的，所以它们作为宏突出显示。@AndreyT我将尝试使用单指针，看看它有什么作用。@Chris我用一组描述实体移动行为的枚举将它分组或者，我们可以看到调用MOVE_TO的周围代码，包括用于存储正方形和实体的变量的全部范围吗？