Are C strings guaranteed to be arrays?

Question

Are C strings (as opposed to std::string) guaranteed to be implemented as arrays? For example, say, I have

char const * str = "abc";

What it boils down to is whether or not str + 4 a legal pointer value (without dereferencing that is). I'm asking this because I dont know if C strings are a special case due to the null character terminating it.

Answer 1

First part of the question

Are C strings guaranteed to be implemented as arrays?

For example, say, I have: char const * str = "abc"

Yes, a string object is of an array type. A character string is a data format and a (character) string object is of a type array of char.

In your example str points to the string literal "abc". Character string literals have the type char[N+1] where N is the length of the string (i.e., the number of characters excluding the terminating null character).

Some references from Standard and K&R 2nd edition:

C defines a string literal as:

(C99, 6.4.5p2) "A character string literal is a sequence of zero or more multibyte characters enclosed in double-quotes, as in "xyz"."

and says (emphasis mine):

C99, 6.4.5p5) "For character string literals, the array elements have type char, and are initialized with the individual bytes of the multibyte character sequence;"

K&R 2nd edition says:

"Technically, a string constant is an array of characters"

and

"when a string constant like "hello\n" appears in a C program, it is stored as an array of characters containing the characters in the string and terminated with a '\0' to mark the end."

Second part of the question

What it boils down to is whether or not str + 4 a legal pointer value (without dereferencing that is).

Yes, it is a valid pointer. In your case str + 4 is a pointer one past the last element of the array.

A valid pointer is a pointer that is either a null pointer or a pointer to a valid object. For an element of an array object, a pointer one past the last element of the array object is also a valid pointer.

Note that for the purpose of the last rule ("the one past element"), for pointers to objects that are not elements of an array, C treats the object as an array of one element.

(C99, 6.5.6p7) "For the purposes of these operators, a pointer to an object that is not an element of an array behaves the same as a pointer to the first element of an array of length one with the type of the object as its element type."

Answer 2

They are guaranteed to be a contiguous sequence of chars. If that's your definition of an array, then yes.

In your example you will have 4 chars, one for each character and one for the null terminator. str+4 will be out of range.