Is char** a pointer to an array of strings in C?

Question

From my understanding, char* X is a variable that points to a single character or a character array (string) in C.

char** is a pointer which points to another pointer which finally points to a single character or a character array.

if int** is equivalent to creating a multidimensional array, why can't I create an array of strings in C using char**?

const char** day = {
    "Sunday",
    "Monday",
    "Tuesday",
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday"
};

Here *day would point to the array itself and **day would point to the first element of the array "Sunday"?

Possible duplicate of [What is double star (eg. NSError \*\*)?](https://stackoverflow.com/questions/604099/what-is-double-star-eg-nserror) — Ronaldo Araújo Alves, Apr 30 '19 at 02:46
@RonaldoAraújoAlves let's not confuse ppl with objective-c here — Antti Haapala -- Слава Україні, Apr 30 '19 at 02:59

Achal · Answer 1 · 2019-04-30T03:01:31.360

why can't I create an array of strings in C using char**? yes you can create. Correct way is to use array of char pointers not the double char** pointer as you did. For e.g

const char* day[7] = {
    "Sunday",
    "Monday",
    "Tuesday",
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday"
};

Above, day is array of char pointers and its each element is character pointer and each elements needs to point to valid address. Here day[0] is character pointer & its pointing to string literal sunday which is valid address.

Its also possible via char** but not the way as you did. For e.g

char **day = {"sunday", "Monday" };

here problem is you haven't allocated memory for day[0], day[1] to hold some string literal like sunday & so on.

First allocate memory for day like below

char **day = malloc(NUM_OF_DAYS * sizeof(*day)); /* define NUM_OF_DAYS as 7 */

And then allocate memory for day[0], day[1] etc.

for(int index =0 ; index < NUM_OF_DAYS; index++) {
   day[index] = malloc(MAX_DAY_SIZE * sizeof(**day)); /* define this MACRO */
}

And then you scan the data at run time.

"here problem is you haven't allocated memory for day[0], day[1]" No it isn't. The string literals are already allocated in read-only memory. The problem is that a `char**` can only be assigned to point at a single `char*`. A `char**` is a _scalar_ (single item) not an _aggregate_ (container), so you can't initialize it like that with `{ a, b }`. It's a syntax error and not related to dynamic allocation. You can however initialize it to point at a compound literal array: `const char** day = (const char* []){"Sunday", "Monday"};`. You only need to malloc if you need read/write memory. — Lundin, Dec 13 '22 at 09:14

score 3 · Accepted Answer · answered Apr 30 '19 at 06:58

char* X is a variable that points to a single character or a character array (string) in C.

No, it is a variable that contains the address of a single character. That single character may or may not be the beginning of a null terminated string.

Thus char** points at a char* that points at a character.

if int** is equivalent to creating a multidimensional array

It is not. int** has nothing to do with multi-dimensional arrays.

why can't I create an array of strings in C using char**?

Because a char** is not an array nor can it, strictly speaking, point at one.

A char* can be used to point at a null terminated string. So if you want an array of strings, you need an array of char*. Which can for example be written as:

char* day[] = { ... };

Now as it turns out, a char** can be used to point at the address of the first item of this array, &day[0], since that is a char*. This often make people confuse char** for multi-dimensional arrays, because it is valid to do this:

char* day[] = { "Sunday", ... };
char** week = &day[0];
printf("%c", week[0][0]); // prints 'S'

But that does not make char** an array, nor does it make it a 2D array.

You can also use char** to dynamically allocate a look-up table of strings, where each string has a variable length:

char** week = malloc(7 * sizeof(char*));
for(int i=0; i<7; i++)
{
  week[i] = malloc( ... );
}

Here again, the char** points at the first item of a 1D array of char*. It is not a 2D array, it does not point at one. That it allows week[i][j] syntax is irrelevant, it is still not an array (and [] is never actually used on array types, Do pointers support “array style indexing”?).

More info: Correctly allocating multi-dimensional arrays.

Is char** a pointer to an array of strings in C?

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