Dynamically allocating array explain

Question

This is sample code my teacher showed us about \"How to dynamically allocate an array in C?\". But I don\'t fully understand this. Here is the code:

int k;
i         


        

Answer 1

Remember that arrays decays to pointers, and can be used as pointers. And that pointers can be used as arrays. In fact, indexing an array can be seen as a form or pointer arithmetics. For example

int a[3] = { 1, 2, 3 };  /* Define and initialize an array */
printf("a[1] = %d\n", a[1]);  /* Use array indexing */
printf("*(a + 1) = %d\n", *(a + 1));  /* Use pointer arithmetic */

Both outputs above will print the second (index 1) item in the array.

The same way is true about pointers, they can be used with pointer arithmetic, or used with array indexing.

From the above, you can think of a pointer-to-pointer-to.type as an array-of-arrays-of-type. But that's not the whole truth, as they are stored differently in memory. So you can not pass an array-of-arrays as argument to a function which expects a pointer-to-pointer. You can however, after you initialized it, use a pointer-to-pointer with array indexing like normal pointers.