'\0' and printf() in C

In C all literal strings are really arrays of characters, which include the null-terminator.

However, the null terminator is not counted in the length of a string (literal or not), and it's not printed. Printing stops when the null terminator is found.

printf returns the number of the characters printed. '\0' is not printed - it just signals that the are no more chars in this string. It is not counted towards the string length as well

int main()
{
    char string[] = "hello";

    printf("szieof(string) = %zu, strlen(string) = %zu\n", sizeof(string), strlen(string));
}

https://godbolt.org/z/wYn33e

sizeof(string) = 6, strlen(string) = 5

Your assumption is wrong. Your string indeed ends with a \0.

It contains of 5 characters h, e, l, l, o and the 0 character.

What the "inner" print() call outputs is the number of characters that were printed, and that's 5.

The null byte marks the end of a string. It isn't counted in the length of the string and isn't printed when a string is printed with printf. Basically, the null byte tells functions that do string manipulation when to stop.

Where you will see a difference is if you create a char array initialized with a string. Using the sizeof operator will reflect the size of the array including the null byte. For example:

char str[] = "hello";
printf("len=%zu\n", strlen(str));     // prints 5
printf("size=%zu\n", sizeof(str));    // prints 6