Question

Why does printing a null char (‘\0’, 0) with %s prints the “(null)” string actually?

Like this code:

char null_byte = '\0';
printf("null_byte: %s\n", null_byte);

…printing:

null_byte: (null)

…and it even runs without errors under Valgrind, all I get is the compiler warning warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat] (note: I’m using gcc 4.6.3 on 32bit Ubuntu)

Answer 1

It’s undefined behavior, but it happens that on your implementation:

• the int value of 0 that you pass is read by %s as a null pointer
• the handling of %s by printf has special-case code to identify a null pointer and print (null).

Neither of those is required by the standard. The part that is required[*], is that a char used in varargs is passed as an int.

[*] Well, it’s required given that on your implementation all values of char can be represented as int. If you were on some funny implementation where char is unsigned and the same width as int, it would be passed as unsigned int. I think that funny implementation would conform to the standard.