Question

I have a string I composed using memcpy() that (when expanded) looks like this:

char* str = "AAAA\x00\x00\x00...\x11\x11\x11\x11\x00\x00...";

I would like to print every character in the string, and if the character is null, print out "(null)" as a substitute for ‘\0’.

If I use a function like puts() or printf() it will just end at the first null and print out

AAAA

So how can I get it to print out the actual word “(null)” without it interpreting it as the end of the string?

Answer 1

You have to do that mapping yourself. If you want to, that is. In C, strings are null-terminated. So, if you use a formatted output function such as printf or puts and ask it to print a string (via the format specifier %s) it’d stop printing str as soon as it hits the first null. There is no null word in C. If you know exactly how many characters you have in str you might as well loop over them and print the characters out individually, substituting the 0 by your chosen mnemonic.

The draft says 7.21.6.1/8:

p The argument shall be a pointer to void. The value of the pointer is
converted to a sequence of printing characters, in an
implementation-defined manner.

However, the following:

$ cat null.c
#include <stdio.h>
int main() {
 printf("%p\n", (void *)0);
}

produces:

00000000

on both gcc 4.6 and clang 3.2.

However, on digging deeper:

$ cat null.c
#include <stdio.h>
int main() {
 printf("%s\n", (void *)0);
}

does indeed produce the desired output:

(null)

on both gcc and clang.

Note that the standard does not mandate this:

s If no l length modifier is present, the argument shall be a pointer
to the initial element of an array of character type.280) Characters
from the array are written up to (but not including) the terminating
null character. If the precision is specified, no more than that many
bytes are written. If the precision is not specified or is greater
than the size of the array, the array shall contain a null character.

Relying on this behavior may lead to surprises!