c - char *array and char array[]

Question

if I write this

 char *array = "One good thing about music";

I actually create an array? I mean it's the same like this?

char array[] = {"One", "good", "thing", "about", "music"};

Answer 1

The declaration and initialization

char *array = "One good thing about music";

declares a pointer array and make it point to a constant array of 31 characters.

The declaration and initialization

char array[] = "One, good, thing, about, music";

declares an array of characters, containing 31 characters.

And yes, the size of the arrays is 31, as it includes the terminating '' character.

---

Laid out in memory, it will be something like this for the first:

+-------+     +------------------------------+
| array | --> | "One good thing about music" |
+-------+     +------------------------------+

And like this for the second:

+------------------------------+
| "One good thing about music" |
+------------------------------+

---

Arrays decays to pointers to the first element of an array. If you have an array like

char array[] = "One, good, thing, about, music";

then using plain array when a pointer is expected, it's the same as &array[0].

That mean that when you, for example, pass an array as an argument to a function it will be passed as a pointer.

Pointers and arrays are almost interchangeable. You can not, for example, use sizeof(pointer) because that returns the size of the actual pointer and not what it points to. Also when you do e.g. &pointer you get the address of the pointer, but &array returns a pointer to the array. It should be noted that &array is very different from array (or its equivalent &array[0]). While both &array and &array[0] point to the same location, the types are different. Using the arrat above, &array is of type char (*)[31], while &array[0] is of type char *.

---

For more fun: As many knows, it's possible to use array indexing when accessing a pointer. But because arrays decays to pointers it's possible to use some pointer arithmetic with arrays.

For example:

char array[] = "Foobar";  /* Declare an array of 7 characters */

With the above, you can access the fourth element (the 'b' character) using either

array[3]

or

*(array + 3)

And because addition is commutative, the last can also be expressed as

*(3 + array)

which leads to the fun syntax

3[array]