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

by

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;

More Related Contents:

  1. Finding target sum of two numbers in array
  2. Pointer manipulation inside function
  3. Please explain the following parallel code template
  4. Why is this string reversal C code causing a segmentation fault? [duplicate]
  5. What can be the reasons of connection refused errors?
  6. C Minishell Adding Pipelines
  7. What does space in scanf mean? [duplicate]
  8. Implicit int return value of C function
  9. Returning a C string from a function
  10. When and why to use malloc?
  11. printf not printing on console
  12. How to call execl() in C with the proper arguments?
  13. Increasing limit of FD_SETSIZE and select
  14. What is the valid range for program return value in Linux/bash? [duplicate]
  15. Generating a random bit – lack of randomness in C rand()
  16. Output of multiple post and pre increments in one statement [duplicate]
  17. how could I intercept linux sys calls?
  18. Why am I able to perform floating point operations inside a Linux kernel module?
  19. understanding C namespaces
  20. Waiting for all child processes before parent resumes execution UNIX
  21. Are “malloc(sizeof(struct a *))” and “malloc(sizeof(struct a))” the same?
  22. What is double star (eg. NSError **)?
  23. Find the Size of integer array received as an argument to a function in c [duplicate]
  24. Deleting a middle node from a single linked list when pointer to the previous node is not available
  25. How to enable linking floating point library in TurboC?
  26. How does srand relate to rand function?
  27. Is calloc(4, 6) the same as calloc(6, 4)?
  28. Is there any option to switch between C99 and C11 C standards in Visual Studio?
  29. Why am i getting this warning in “if (fd=fopen(fileName,”r”) == NULL)”?
  30. What is the function of an asterisk before a function name?

Leave a Comment