数组在内存中是如何存储和删除的？

Question

I made a 2D array on the heap of some objects:

Step (1)

Obj **arr = new Obj *[n];

for (int i=0;i<n;i++)
{
    arr[i] = new Obj[n];
}

// so this creates 2D arr[n][n]...then when I delete it:

Step (2)

for (int i=0;i<n;i++)
{
    delete [] arr[i];
}
delete [] arr;

So I'm still not sure what this delete does. Does it run the destructor of Obj and flag the OS telling it this is now available memory.

Now what I REALLY do not understand is that when I do Step (1) again (after I deleted), I get these objects initialized to weird values, yet this doesn't happen the first time I do it (all zero-initialized). Did I just get lucky the first time?

Answer 1

Your example lacks the declaration of Obj.

• new[] allocates memory and calls the constructor of each element
• If the constructor does not alter memory, you will see some random values - maybe zeros.
• delete[] calls the destructor of each element previously allocated with new[] and deallocates the memory, finally.
• In a debugging compilation the memory might be filled with some bytes indicating the deallocation.
• Doing new[] right after the deallocation might show indicator bytes.

Answer 2

AFAIK, the following code will NOT give you weird values, no matter how many times you repeat deleting and newing.

#include <iostream>
using namespace std;

class Foo
{
public:
    Foo(): val(-2) { cout << "ctor" << endl; }
    ~Foo() { cout << "dtor: " << val << endl; }

private:
    int val;
};

int main()
{
    Foo **arr = new Foo *[2];
    for (int i = 0; i < 2; ++i)
        arr[i] = new Foo[2]();    // <- for builtin type, () is required to initialized to zero.

    for (int i = 0; i < 2; ++i)
        delete [] arr[i];
    delete [] arr;
    return 0;
}

Relevant post: Operator new initializes memory to zero

As to what happens to pointers after you delete them, please see this post: C - What happens to an array of pointers when the array is freed?