passing 2D array pointer to function in C - c

Question

I'm trying to pass the address of a 2D array to a function in C. I initialize the 2D array as:
const int N = 3;
char t[N][N];
I try to convert this into a char***:
char*** t_ptr = &t;
But it fails with a:
warning: initialization from incompatible pointer type
The function I want to receive the array has a prototype such as:
void f(char*** t, int N) { ... };
What am I doing wrong? Thank you.

Answer 1

This
char*** t_ptr = &t;
is wrong as compiler pointed because t is an two dimensional array, instead of triple pointer like char*** use pointer to an array to point to it. For e.g
char t[3][3] = { "ab","de","gh"}; /* total 3 1D array & each 1D array has 3 elements */
char (*t_ptr)[3] = t; /* t_ptr is a pointer to an array, pointing to 3 elements at a time */
And you can print t_ptr like
for(int index = 0; index < 3 ; index++) {
printf("%s\n",t_ptr[index]);
}

Answer 2

For reading c declarations, you can visit: https://cdecl.org/
Now using #Achal's example array:
#include <stdio.h>
#include <stdlib.h>
int main()
{
// t is a 2D array (not a VLA)
char t[3][3] =
{
{'a', 'b', '\0'},
{'d', 'e', '\0'},
{'g', 'h', '\0'}
};
printf("t is composed of 3 arrays of 3 characters each,\n");
printf("with the following addresses (on my machine):\n");
printf("--------------------------------------------------\n");
printf("%p, %p, %p\n", t[0], t[1], t[2]);
// ------------------------------------------------
// p is an array of 3 char pointers or char*
// Notice initialization
char* p[3] = {t[0], t[1], t[2]};
// following line of code will break, if no '\0' is encountered
// since %s requires a char* and a null terminator
printf("\nPrint strings using p:\n");
printf("------------------------\n");
printf("%s, %s, %s\n", *p, *(p + 1), *(p + 2));
// ------------------------------------------------
// q is a pointer to a char* or q has type char**
// Notice initialization of q (q points to p)
char** q = p;
printf("\nUsing q:\n");
printf("-----------\n");
printf("%s, %s, %s\n", *q, *(q + 1), *(q + 2));
// ---------------- printing characters individually
printf("\nIndividually:\n");
printf("---------------\n");
for(int i = 0; i < 2; i++)
{
for(int j = 0; j < 2; j++)
{
printf("%c ",
*(*(q + i) + j)
);
}
printf("\n");
}
// -----------------------------------------------
// r is a pointer to an array of size 3 containing char
// r advances three char at a time (r doesn't know the size of t)
char (*r)[3] = t; // this is the type of t
printf("\nUsing r:\n");
printf("---------------\n");
printf("%p, %p, %p\n", *r, *(r + 1), *(r + 2));
// to extract chars
printf("%c, %c", *(*(r + 0) + 0), *(*(r + 2) + 1));
// -----------------------------------------------
return EXIT_SUCCESS;
}
Output:
t is composed of 3 arrays of 3 characters each,
with the following addresses (on my machine):
--------------------------------------------------
000000000022FE2F, 000000000022FE32, 000000000022FE35
Print strings using p:
------------------------
ab, de, gh
Using q:
-----------
ab, de, gh
Individually:
---------------
a b
d e
Using r:
---------------
000000000022FE2F, 000000000022FE32, 000000000022FE35
a, h
Process returned 0 (0x0) execution time : -0.000 s
Press any key to continue.

Answer 3

char t[N][N];
is in fact the same as
char t[N * N];
in memory. A pointer to such an array would in both cases be of type char *.
char *** is a pointer to a pointer, that is a pointer to a char, whereas char * is a pointer to a char and that's how you pass array references in C: You pass them as a pointer to the first element of that array and this first element is a char.
C cannot retain the exact type or structure of an array as soon as you pass it to functions. In memory, a char array is just a bunch of memory filled with chars and all you can pass around is a pointer to that memory. If that memory is char [] or char [][] or even char [][][] plays no role, in memory all three are a block full of chars and the function would have to explicitly know the structure in memory, otherwise all char arrays will always be char [] for a function.
I strongly discourage C beginners to ever use multidimensional arrays. Instead of
char t[N][N];
char c = t[y1][x1];
t[y2][x2] = 'X';
use
char t[N];
char c = t[y1 * N + x1];
t[y2 * N + x2] = 'X';
As that's basically what the compiler will internally do anyway.
Note that multidimensional arrays in C are not x-y, but y-x, the first value is the row, the second on the column, please see this tutorial.
Whoever disbelieves what I just said, try out this code:
int main ( ) {
char a[5][5];
for (int y = 0; y < 5; y++) {
for (int x = 0; x < 5; x++) {
a[y][x] = x + 10 * y;
}
}
for (int y = 0; y < 5; y++) {
for (int x = 0; x < 5; x++) {
printf("%02d ", a[y][x]);
}
printf("\n");
}
printf("------\n");
char * a2 = (char *)a;
for (int y = 0; y < 5; y++) {
for (int x = 0; x < 5; x++) {
printf("%02d ", a2[y * 5 + x]);
}
printf("\n");
}
}
You can run it online, if you like, the output is identical. Also have a look at the assembly code the compiler generates for either loop (e.g. gcc -S) and you will see it's almost identical as even in the first case the compiler uses an add and a mul instruction to access the correct memory location within the array.