＆＃34;删除＆＃34;运算符C ++

Question

我理解的是，如果您使用＆＃34;删除＆＃34;指向堆上的内存地址的指针上的运算符，它释放内存。我编写了这段代码来测试我的知识，但即使使用了delete操作符，内存地址仍然具有地址的值。有谁能解释一下这里发生了什么？

#include <iostream>

using namespace std;

int main()
{
   int* aPtr = new int(5);
   cout << "address of int in heap is " << aPtr << endl;
   cout << "value of int in heap is " << *aPtr << endl;
   delete aPtr;
   cout << "address of int in heap is " << aPtr << endl;
   cout << "value of int in heap is " << *aPtr << endl;
   return 0;
}

我得到的控制台输出在这里，

address of int in heap is 0x7fe851402700
value of int in heap is 5
address of int in heap is 0x7fe851402700
value of int in heap is 5