Why does printf literally print (null) and what exactly happens?

Question

in a C programming exercise I'm doing something like this (just simplifying):

printf( "%s", 0);

The output is

(null)

What happens here? I assume that printf interprets the zero as a char *, so to NULL? How could I replicate this result by something like

char string[] = NULL; //compiler-error
printf( "%s", string);

?

Answer 1

Firstly, your

printf("%s", 0);

leads to undefined behavior (UB). %s in printf requires a char * pointer as argument. You are passing 0, which is an int. That alone already breaks your code, just like

printf("%s", 42); 

would. For that specific UB the fact that 0 is a zero does not make any difference.

Secondly, if you really want to attempt to pass a null-ponter to %s format specifier, you have to do something like

printf("%s", (char *) 0);

Of course, this leads to undefined behavior as well, since %s requires a pointer to a valid string as argument, and (char *) 0 is not a valid string pointer. But some implementations prefer to handle such situations gracefully and just print (null).

In your particular case you just got lucky: printf("%s", 0) "worked" the same way as printf("%s", (char *) 0) would and your implementation saved the day by outputting (null).

Answer 2

As others have noted, passing a null pointer to printf %s is not guaranteed to do anything. Everything else being equal, we would expect a segmentation violation or other ungraceful crash, as printf attempts to dereference the null pointer. As a convenience, however, many (most?) implementations of printf have, somewhere deep within them, the equivalent of

case 's':
    char *p = va_arg(argp, char *);
    if(p == NULL) p = "(null)";
    fputs(p, stdout);