使用new分配内存返回相同的内存地址

问题描述：

class abc  {
    int x ;
};
int main()
{
    abc *A = new abc ;
    cout<< static_cast<void*>(A) << endl ;
    delete A ;
    cout<< static_cast<void*>(A) << endl ;
    abc *B = new abc ;
    cout<< static_cast<void*>(B) << endl ;
    delete B ;
    cout<< static_cast<void*>(B) << endl ;



    return 0;

}

即使我删除了旧内存，为什么它仍打印相同的地址.
即使我在删除后指定null，它也会打印相同的地址.

Why it prints the same address , even though i deleted old memory .
Even i assign null after deletion , it prints the same address.

A和B的地址都相同.

Even address of A and B is same .

实时场景:

Function1 ->   
arr[p] = new X  
ptr1 = arr[p]  
using ptr1 

Function2 ->   
ptr2 = arr[p]  
delete ptr2  
arr[p] = new X ( new data)

在实际情况下，应该使ptr1无效，但是由于编译器将相同的地址分配给
function2 ptr1中的arr [p]仍然有效.

In real scenario ptr1 should be invalidated but since compiler assign same address to
arr[p] in function2 ptr1 still works .

答

为什么不应该这样呢?一旦删除了特定地址的内存，内存管理器便可以自由地在下次请求新内存时重用该地址.确实，这是内存管理器使用的非常常见的优化.他们会跟踪最近释放的块，并将其交给要求该大小的块的下一个客户端.

Why should this not happen? Once you've deleted memory at a particular address, the memory manager is perfectly at liberty to re-use that address the next time you ask for new memory. Indeed this is a very common optimisation used by memory managers. They keep track of recently freed blocks and hand them back to the next client that requests a block of that size.

查看此问题的另一种方法是考虑如果从未重新使用释放的地址，将会发生什么情况.如果那要发生，那么在足够的分配/取消分配周期之后，最终将没有剩余的地址空间.确实，如果重用从未发生过，那么分配内存将毫无意义.因此，可以，请期望当您释放内存时，该内存地址将被重用.

Another way to look at this would be to consider what would happen if freed addresses were never re-used. If that were to happen then eventually, after enough allocation/deallocation cycles, there would be no address space left. Indeed, if re-use never happened then there would be no point at all in deallocating memory. So yes, do expect that when you deallocate memory, that memory address will be re-used.

使用new分配内存返回相同的内存地址

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