c++ - 在堆上创建的变量，指向同一个变量的2个指针有不同的地址？

Question

刚学了栈和堆的区别。在创建一个将为我在堆上动态分配内存的函数后，我返回指针并显示(在函数内和函数外)每个指针的地址和值。

值是相同的，这是我所期望的，但是堆上同一 block 内存的地址是不同的，这是我没有想到的。

为什么？ pHeap2 和 pTemp 不应该指向同一个地址吗？

#include <iostream>
using namespace std;

int* intOnHeap(); // returns an int on the heap
int main()
{
 int* pHeap = new int; // new operator allocates memory on the heap and returns its address
       // 'new int' allocates enough memory on heap for one int and returns the address on the heap for that chunk of memory
       // 'int* pHeap' is a local pointer which points to the newly allocated chunk of memory
 *pHeap = 10;
 cout << "*pHeap: " << *pHeap << "\n\n";

 int* pHeap2 = intOnHeap();
 cout << "pHeap2:\n-----------" << endl;
 cout << "Address:\t" << &pHeap2 << "\n";
 cout << "Value:\t\t" << *pHeap2 << "\n\n";

 cout << "Freeing memory pointed to by pHeap.\n\n";
 delete pHeap;

 cout << "Freeing memory pointed to by pHeap2.\n\n";
 delete pHeap2;

// get rid of dangling pointers
    pHeap = 0;
    pHeap2 = 0;

 system("pause");
 return 0;
}

int* intOnHeap()
{
 int* pTemp = new int(20);
 cout << "pTemp:\n-----------" << endl;
 cout << "Address:\t" << &pTemp << "\n";
 cout << "Value:\t\t" << *pTemp << "\n\n";
 return pTemp;
}

输出:

*pHeap: 10

pTemp:
-----------
Address:        0042FBB0
Value:          20

pHeap2:
-----------
Address:        0042FCB4
Value:          20

Freeing memory pointed to by pHeap.

Freeing memory pointed to by pHeap2.

Press any key to continue . . .

Answer 1

您报告的是指针的地址，而不是指针指向的地址。当然，pTemp 和 pHeap2 的指针地址会有所不同；这些是恰好指向内存中相同地址的不同指针。删除 & 前缀 pTemp 和 pHeap2 以查看您期望的结果。

图片是这样的:

0042FBB0           0042FBC0     0042FCB4
------------       ------       ------------
| 0042FBC0 |------>| 20 |<------| 0042FBC0 |
------------       ------       ------------
pTemp                           pHeap2

此处 &pTemp 为 0042FBB0，&pHeap2 为 0042FCB4。我为 pTemp 和 pHeap2 指向的地址创建了一个地址(当然，结果可能因运行而异)并且如果您删除 & 前缀 pTemp 和 pHeap2 然后你会看到 0042FBC0 在每种情况下打印出来。