C++ 无法删除二维数组的第二维

Question

I am trying to make an array of pointers and each pointer to be an array of only 1 integer. Then i want to print this integer and after i print it, i want to delete this pointer so that it can not be printed again. The problem is that this command "delete[] id_number[i];" seems not to work.

The code:


using namespace std;

int main()
{
    int** id_number= new int* [12];
    for (int i = 0; i < 12; i++)
    {
        id_number[i] = new int[1];
        id_number[i][0] = i;
    }

    for (int i = 0; i < 12; i++)
    {
        cout << "id_number ["<< i << "] [0] = "<< id_number[i][0] << endl;

        delete[] id_number[i];        //Trying to delete each pointer after printing the 
                                        //integer         
    }

    cout << endl << "After the end of loop" << endl;
    cout << "id_number [" << 5 << "] [0] = " << id_number[5][0] << endl;

    return 0;


The outcome:


id_number [0] [0] = 0
id_number [1] [0] = 1
id_number [2] [0] = 2
id_number [3] [0] = 3
id_number [4] [0] = 4
id_number [5] [0] = 5
id_number [6] [0] = 6
id_number [7] [0] = 7
id_number [8] [0] = 8
id_number [9] [0] = 9
id_number [10] [0] = 10
id_number [11] [0] = 11

After the end of loop
id_number [5] [0] = 5     // why does this print work? didn't i delete the pointer "id_number 
                            // [5]" inside the loop??

Answer 1

After deleting the entries with

delete[] id_number[i]

set the pointer you just deleted to a nullptr like this

delete[] id_number[i]; id_number[i] = nullptr;

After deleting the alle the pointer in the arry you still have your array which occupies recources. therefore it also has to deleted

delete[] id_number; id_number = nullptr;

This should free all the resources correctly. If you then as mentioned by WhozCraig try to acces deleted resources illegaly, undefiend behvoir will occure and my look like it does would it should which is again as mentioned by WhozCraig just undefiend behavoir.