Question

There is such code:

int tab[14][2];
int (*wskk)[2] = tab; // &tab makes error

int tab2[2];
wskk = &tab2; // tab2 makes error

Why is it possible to use one pointer to point at two arrays of different dimensions?

Answer 1

To understand what’s going on you must be familiar with a few key-concepts:

• a multidimensional array is an array of arrays;
• the name of an array decays to a pointer to its first element;
• the type of wskk is “pointer to an array of 2 ints”.

Thus, if you write tab you’re getting a pointer to the first element of tab, which is its first row; the row has type int[2], so a pointer to it has type int (*)[2], which is exactly the type of your pointer. Because of this you can assign tab to wskk, which will now point to the first row of tab.

You can’t assign &tab to it, because that yields you a pointer to the whole multidimensional array, which is of type int (*)[14][2].

As for the second piece, it’s even simpler: tab2 is an array of two ints, so its type is int[2]. If you get a pointer to it via the & operator, you get a int (*)[2], which is the type of your pointer. Actually, it makes sense: tab2 and a row of tab are effectively the same stuff (an array of 2 ints).

You can’t assign tab2 to it because tab2 decays to a pointer to its first element, i.e. an int *.