C++ 無法刪除二維數組的第二維

Question

我正在嘗試創建一個指針數組，每個指針都是一個只有 1 個 integer 的數組。 然后我想打印這個 integer 並且在我打印它之后，我想刪除這個指針，這樣它就不能再打印了。 問題是這個命令“delete[] id_number[i];” 似乎不起作用。

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

刪除條目后

delete[] id_number[i]

將剛剛刪除的指針設置為這樣的 nullptr

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

刪除數組中的所有指針后，您仍然擁有占用資源的數組。 因此它也必須刪除

delete[] id_number; id_number = nullptr;

這應該正確釋放所有資源。 如果您然后如 WhozCraig 所述嘗試非法訪問已刪除的資源，則會發生不安全行為，而我看起來確實應該這樣做，正如 WhozCraig 再次提到的那樣，只是不安全行為。