当指向动态分配的内存时，指针的指针将分配什么值？

Question

Consider the following case:

int **my_array = new int*[10];

1. What do we assign to my_array here?
2. my_array is a pointer that points to what?
3. Is there any way to iterate through my_array (the pointer) and set up a two-dimensional array of integers (and not int*)?

Answer 1

We assign to my_array the address of an array. The array contains pointers which can point to other arrays, but don't yet.

Yes, we can do this:

int **my_array = new int*[10];

for(int i=0; i<10; ++i)
  my_array[i] = new int[13];

my_array[2][11] = 500;

Answer 2

What do we assign to my_array here?

You can assign an int* to the elements of my_array . Eg

my_array[0] = new int[20];

or

int i;
my_array[0] = &i;

my_array is a pointer that points to what?

It points to an an array of 10 int* objects.

Is there any way to iterate through my_array (the pointer) and set up a two-dimensional array of integers (and not int*)?

Not sure what you are expecting to see here. An element of my_array can only be an int* .

If you want my_array to be a pointer to a 2D array, you may use:

int (*my_array)[20] = new int[10][20];

Now you can use my_array[0][0] through my_array[9][19] .

PS If this is your attempt to understand pointers and arrays, it's all good. If you are trying to deploy this code in a working program, don't use raw arrays any more. Use std::vector or std::array .

For a 1D array, use:

// A 1D array with 10 elements.
std::vector<int> arr1(10);

For a 2D array, use:

// A 2D array with 10x20 elements.
std::vector<std::vector<int>> arr2(10, std::vector<int>(20));