我是一名大学生，我正在尝试解决教授给我的C ++练习。我的英语不好意思。简而言之，我必须不使用数组而是仅使用指针来管理一定数量（相同类型）的对象。有问题的对象是“房屋”类型的对象。
我创建了指向“房屋”的指针，并通过循环将指针指向了由用户输入的变量初始化的“房屋”类型的新对象。然后，我滑动指针并重新开始。
House* housePtr;
cout<<"We start building houses, you will have to build 4.\n";
for (auto i=0; i<4; ++i, ++housePtr)
{
int r, d;
cout<<"\nLet's build the number "<<i+1<<endl;
cout<<"How many rooms must it have?\n";
cin>>r;
cout<<"\nHow far is it from the center?\n";
cin>>d;
housePtr= new House(r, d);
}
问题出现在我滚动指针所指向的对象时。例如打印我创建的对象拥有的值。
先前的for循环使我指向了最后创建的对象旁边的位置的指针。因此，通过for循环，我将指针带回到第一个对象（因此使指针向后退四步），并且在每次迭代时，我让他打印指针所持有的内存地址，即每个House的内存地址。
for (auto i=0; i<4; i++, housePtr--)
{
cout<<endl<<housePtr<<endl;
}
这是最后一段代码的输出：
0x10139c
0x101390
0x101384
0x101378
第一个是与对象无关的地址，因为它是最后一个对象之后的位置所固有的地址。按照我的逻辑，后面的其他3个分别是第四，第三和第二宫的地址。
再次使用指针，对于每个对象，我都打印了它的值和地址
for (auto i=0; i<4; housePtr++, i++)
{
cout<<"\nThe house "<<i+1<<" has "<<housePtr->getNumOfRooms()<<" rooms and is ";
cout<<housePtr->getDistanceFromCenter()<<" meters from the center\n";
cout<<housePtr<<endl;
}
这是输出：
The house 1 has 190607135 rooms and is 201338508 meters from the center
0x10136c
The house 2 has 7 rooms and is 4 meters from the center
0x101378
The house 3 has 190607135 rooms and is 201338508 meters from the center
0x101384
The house 4 has 5 rooms and is 8 meters from the center
0x101390
我输入的初始输入是：
We start building houses, you will have to build 4.
Let's build the number 1
How many rooms must it have?
8
How far is it from the center?
7
Let's build the number 2
How many rooms must it have?
5
How far is it from the center?
8
Let's build the number 3
How many rooms must it have?
7
How far is it from the center?
4
Let's build the number 4
How many rooms must it have?
5
How far is it from the center?
8
我不明白为什么它不能正确打印数据，以及由于什么原因在一次迭代中它打印对象的数据，而在下一次它打印随机数。
问题出在哪里？
这里，当您进行housePtr = new House（r，d）;时，将创建新对象，并且housePtr指向该对象。现在，当您执行housePtr ++时，它将按House类的大小递增。现在再次执行housePtr = new House（r，d）;时，将创建新对象，并且不一定是先前创建对象的连续地址。这是C ++中最重要的事情。 C ++为您提供了赤裸裸的记忆，无视安全性。蜘蛛侠说这是巨大的力量，“力量越大，责任越大”。在这里，您始终必须确保在动态创建对象时，始终存储指向该对象的指针。否则，您将永远无法获得该地址。这确实是一个巨大的问题，名为memory-leak。
只是一个建议，我想您的教授希望您使用链接列表。如果使用链表，则不必创建某种指针数组来指向所有创建的对象。在链接到所有对象的皮棉地址中，会将其存储在先前的节点指针字段中。
增加不是数组的指针将导致该指针指向未使用/无效的内存，这将导致未定义的行为（这就是为什么您得到奇怪的值）。每次也用new覆盖指针将导致内存泄漏。 Baiscally，您使用的指针就像是是数组一样，即使不是。如果您始终必须制作4个House对象，请按照以下步骤操作：
House* housePtr1;
House* housePtr2;
House* housePtr3;
House* housePtr4;
cout<<"We start building houses, you will have to build 4.\n";
int r, d;
cout<<"\nLet's build the number 1"<<endl;
cout<<"How many rooms must it have?\n";
cin>>r;
cout<<"\nHow far is it from the center?\n";
cin>>d;
housePtr1 = new House(r, d);
cout<<"\nLet's build the number 2"<<endl;
cout<<"How many rooms must it have?\n";
cin>>r;
cout<<"\nHow far is it from the center?\n";
cin>>d;
housePtr2 = new House(r, d);
cout<<"\nLet's build the number 3"<<endl;
cout<<"How many rooms must it have?\n";
cin>>r;
cout<<"\nHow far is it from the center?\n";
cin>>d;
housePtr3 = new House(r, d);
cout<<"\nLet's build the number 4"<<endl;
cout<<"How many rooms must it have?\n";
cin>>r;
cout<<"\nHow far is it from the center?\n";
cin>>d;
housePtr4 = new House(r, d);
然后打印值：
cout<<"\nThe house 1 has "<<housePtr1->getNumOfRooms()<<" rooms and is ";
cout<<housePtr1->getDistanceFromCenter()<<" meters from the center\n";
cout<<housePtr1<<endl;
cout<<"\nThe house 2 has "<<housePtr2->getNumOfRooms()<<" rooms and is ";
cout<<housePtr2->getDistanceFromCenter()<<" meters from the center\n";
cout<<housePtr2<<endl;
cout<<"\nThe house 3 has "<<housePtr3->getNumOfRooms()<<" rooms and is ";
cout<<housePtr3->getDistanceFromCenter()<<" meters from the center\n";
cout<<housePtr3<<endl;
cout<<"\nThe house 4 has "<<housePtr4->getNumOfRooms()<<" rooms and is ";
cout<<housePtr4->getDistanceFromCenter()<<" meters from the center\n";
cout<<housePtr4<<endl;
我猜您的教授想提出的观点是，在没有数组的情况下完成所有这些工作很痛苦;）