C/C++ Array initializer problem

Question

Can someone explain why I can do:

int x[4] = { 0, 1, 2, 3 };
int *p = x;

But cannot do:

int x[2][2] = { 0, 1, 2, 3 };
int **p = x;

Is there a way to be able to assign to **p the x[][]?

TIA

Answer 1

A pointer to int int * can point at an integer within an array of integers, so that's why the first is OK;

int x[4] = { 0, 1, 2, 3 };
int *p = x;

This makes p point to x[0], the first int in x.

A pointer to a pointer-to-int int ** can point at a pointer-to-int within an array of pointers-to-int. However, your second array is not an array of pointers-to-int; it is an array of arrays. There are no pointers-to-int, so there is nothing sensible to point int **p at. You can solve this in two ways. The first is to change the type of p so that it is a pointer-to-array-of-2-ints:

int x[2][2] = { 0, 1, 2, 3 };
int (*p)[2] = x;

Now p points at x[0], the first int [2] in x.

The alternative is to create some pointers-to-int to point at:

int x[2][2] = { 0, 1, 2, 3 };
int *y[2] = { x[0], x[1] };
int **p = y;

Now p points at y[0], which is the first int * in y. y[0] in turn points at x[0][0], and y[1] points at x[1][0].

Answer 2

Using the pointer notation you can do this

*(p + n) will return back a value in the subscript n which is equivalent of p[n], a "single dimension array"

*( * (p + i) + j) will return back a value in the subscripts i and j which is equivalent of p[i][j], a "double dimension array"

This will prove that using array subscripts [] decays into pointers as per the rule - see ANSI C Standard, 6.2.2.1 - "An array name in an expression is treated by the compiler as a pointer to the first element of the array", furthermore, ANSI C Standard 6.3.2.1 - "A subscript is always equivalent to an offset from a pointer", and also ANSI C Standard 6.7.1 - "An array name in the declaration of a function parameter is treated as a pointer to the first element in the array"

If you are not sure in understanding how pointers and arrays work, please see my other answer which will explain how it works...

Please ignore anyone who says arrays and pointers are the same - they are not...