Difference between 2D char Array and char** (OR, 3D char Array and char*** etc)

Question

Firstly, I've already reviewed these:

How are multi-dimensional arrays formatted in memory?

Memory map for a 2D array in C

From there, it is known that a 2D array is not the same as char** but in memory they look exactly same. This sounds strange and so I've researched the following:

#include <stdio.h>

char func(char** m) {
    return m[0][0]; //only works for char**, already discussed in the other SO question
}

int main() {

    //char a[4][2]; //a 2D char array

    int row = 4, col = 2;   //char**
    char** a = malloc(row * sizeof(char*));
    int i;
    for (i = 0; i < row; i++) {
        a[i] = malloc(col * sizeof(char));
    }

    //checking the output
    printf("      &a = %u\n", &a);
    printf("   &a[0] = %u\n", &a[0]);
    printf("&a[0][0] = %u\n", &a[0][0]);
    printf("       a = %u\n", a);
    printf("    a[0] = %u\n", a[0]);

    //printf(" a[0][0] = %u\n", a[0][0]);   //char value in a[0][0], here a garbage value

    //char m = func(a); //only works for char**, already discussed in the other SO question

    return 0;
}

Possible output for char** :

      &a = 3209288  //                  &a
   &a[0] = 4083720  // &(*(a+0))      =  a
&a[0][0] = 4083784  // &(*(*(a+0)+0)) = *a
       a = 4083720  //                   a
    a[0] = 4083784  // *(a+0)         = *a

Possible output for 2D char array :

      &a = 3473104  // &a
   &a[0] = 3473104  //  a
&a[0][0] = 3473104  // *a
       a = 3473104  //  a
    a[0] = 3473104  // *a

It is easy to understand the output of char**. But the output of 2D char array looks strange though it was discussed in the other SO question. I cannot think of a pointer x of whatever data-type when,

x = &x = *x

and all the 3 things physically reside in the same block of memory. Hope that my confusion is understandable. Can anyone explain the mystery?

Answer 1

When you use name of the array is any expression except &array and sizeof array this name will be automaticly converted to pointer to the first element of the array (char [5][10] will be converted to char (*)[10]). Address of this pointer will be equal to the adress of entire array.
So, "sizeof (char [5][10]) == 50", there is no additional pointers.

char arr[5][10];

&a = 3473104 // Address of entire array, (char (*)[5][10])
&a[0] = 3473104 // Address of first row, (char (*)[10])
&a[0][0] = 3473104 // Address of first char, (char *)
a = 3473104 // "a", (char [5][10]), converted to "&a[0]", (char (*)[10])
a[0] = 3473104 // "a[0]", (char[10]), converted ro "&a[0][0]", (char *)