Why can't we use double pointer to represent two dimensional arrays?

Question

Why can\'t we use double pointer to represent two dimensional arrays?

arr[2][5] = {\"hello\",\"hai\"};
**ptr = arr;

Here why doesn\'t the d

Answer 1

I'm going to try to draw how

int array[10][6];

and

int **array2 = malloc(10 * sizeof *array2);
for (int i = 0; i < 10; ++i)
    array2[i] = malloc(6 * sizeof **array2);

look like in the memory and how they are different (And that they can't be cast to each other)

array looks like:

 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
| | | | | | | | | | | | | ..............| | | (10*6 elements of type int)
 - - - - - - - - - - - - - - - - - - - - - -
< first row >< second row> ...

array2 looks like:

 _ _ _ _ _ _ _ _ _ _ 
| | | | | | | | | | | (10 elements of type int *)
 - - - - - - - - - - 
 | |     ....      |     _ _ _ _ _ _
 | |                \-->| | | | | | | (6 elements of type int)
 | |                     - - - - - -
 | |
 | |      _ _ _ _ _ _
 |  \ -->| | | | | | | (6 elements of type int)
 |        - - - - - -
 |
 |
 |      _ _ _ _ _ _
  \ -->| | | | | | | (6 elements of type int)
        - - - - - -

When you say array[x][y], it translates into *((int *)array+x*6+y)

While, when you say array2[x][y], it translates into *(*(array2+x)+y) (Note that for array, this formula also works (read to the end of the post, and then the comments)).

That is, a static 2d array is in fact a 1d array with rows put in one line. The index is calculated by the formula row * number_of_columns_in_one_row + column.

A dynamic 2d array, however is just a 1d array of pointers. Each pointer then is dynamically allocated to point to another 1d array. In truth, that pointer could be anything. Could be NULL, or pointing to a single variable, or pointing to another array. And each of those pointers are set individually, so they can have different natures.

If you need to pass the pointer of array somewhere, you can't cast it to int ** (imagine what would happen. The int values of the cells of array are interpreted as pointers and dereferenced -> Bam! Segmentation fault!). You can however think of array as a 1d array of int [6]s; that is a 1d array of elements, with type int [6]. To write that down, you say

int (*p)[6] = array;