Does C add '\0' when you pass a literal string as an argument in a function?

Question

For instance, if I write:

void function(char *k){ printf("%s",k);}

and call it like this:

function("hello");

does the code translate that string to: "hello\\0" ? Or I'm the one who has to add it?

Answer 1

In C (And C++), when you do const char* mystr = "Hello"; , the compiler will generate the following in (read-only) RAM:

0x7fff2fe0: 'H'
0x7fff2fe1: 'e'
0x7fff2fe2: 'l'
0x7fff2fe3: 'l'
0x7fff2fe4: 'o'
0x7fff2fe5: '\0'

Then, the compiler will replace

const char* mystr = "Hello";

with

const char* mystr = 0x7fff2fe0;

For your usage, your code will turn into

function(0x7fff2fe0)

Simple as that.

On a compiler level, all string literals have type const char[N] , where the char array is an array that contains all of the written characters, followed by a \\0 . The char[N] has a length N that is 1 + the length of the string you write ( char[6] for "Hello" ). More information can be found in the here , where they also use the string "Hello" as an example. Thus, sizeof("Hello") == 6 , and "Hello"[5] == '\\0' (Yes, "Hello"[5] is legal, remember, "Hello" has type const char[6] ). We see this information exemplified in the following:

printf("%d\n", sizeof("Hello")); // 6
const char[] str = "Hello"; // Casts from const char[6] to const char[6]
                            // Resulting in a copy of all 6 bytes
printf("%d\n", sizeof(str)); // 6
const char* str2 = "Hello"; // Casts from const char[6] to const char*
printf("%d\n", sizeof(str2)); // 4 on a 32bit system, 8 on a 64bit system

Do note, when casting to a pointer, that you get some pointer eg 0x7fff2fe0 to an array of characters that is not modifiable - attempting to modify the data pointed at 0x7fff2fe0 or 0x7fff2fe5 is explicitly undefined behavior. This status is commonly represented with const ; by writing const , the compiler will correctly complain if you try to edit it.

As an additional note, by writing

char[] myarr = "Hello";

You will create a duplicate stack-allocated character array named myarr , and that array may be modified. myarr will indeed still contain \\0 and have a size of 6 char s, in particular, myarr will have type char[6] , with sizeof(myarr) == 6 .

Answer 2

From the C11 Standard Section 6.4.5 String Literals , Paragraph 6 (p. 71):

In translation phase 7, a byte or code of value zero is appended to each multibyte character sequence that results from a string literal or literals. 78) The multibyte character sequence is then used to initialize an array of static storage duration and length just sufficient to contain the sequence

Answer 3

A string literal already includes a terminating \\0 by itself, regardless of what you do with that literal. "hello" is always a char [6] array of h , e , l , l , o and \\0 , by definition. So, the fact that you "pass it to a function" is completely inconsequential here.

There's no need to add anything.

Answer 4

String literals are not passed to the functions only the pointer to the first character. The referenced object will have all the chars + terminating zero.