Pointer to pointer and pointer to array disambiguation

I was under the impression that these two were the same, yet, after having problems doing this:

Code:

---

    char **ptr;
    char dd_arr[2][4]; // initialize with something here...
 
    ptr = dd_arr; //compiler error: incompatible pointer types assigning to char **' from 'char [2][4]' [-Wincompatible-pointer-types]

---

I went on a google search that led me to believe they might be different after all.

I'm confused now. *ptr, holds the address of a pointer that holds the address of a char. Similarly, dd_arr[0] holds the address of an area in memory, where the address of a char is stored. I'm not understanding the difference. Why can I do *ptr = *dd_arr, but not ptr = dd_arr?


Thanks.

The 'arrays decay to pointers' rule only applies to the outermost dimension of the array.

So.

Code:

---

char dd_arr[2][4]; // initialize with something here...
char (*ptr)[4];  // ptr to an array of char[4]
ptr = dd_arr;  // or ptr = &dd_arr[0] etc

---

You need the () in the pointer declaration, because char *ptr[4] is an array of pointers to char.