printf with %s to include null characters - c

Question

I need to concatenate some strings, and I need to include NULL bytes. I don't want to treat a '\0' as a terminating byte. I want to save my valuable NULL bytes!
In a code example, if
char *a = "\0hey\0\0";
I need to printf in a format that will output "\0hey\0\0".
-AUstin

Answer 1

How about:
int i;
for(i = 0; i < 4; i++)
printf("%c", a[i]);
If you want a 'printf-like' function to use this when you specify %s in a format string you could include the above code in your own function. But as #Neil mentioned, you'll struggle finding an alternative to looking for null bytes to determine the length of strings. For that I guess you could use some kind of escape character.

Answer 2

The issue here is that the length of the string a cannot be easily determined. For example, your code..
char *a = "\0hey\0\0";
.. allocates seven bytes to the string, the last being the NULL terminator. Using a function like strlen would return 0.
If you know the precise length of the string, then you can write or iterate over the bytes thus:
#ifdef ESCAPE_NULLS
int i;
for (i = 0; i <= 6; i++)
if (a[i] == 0)
printf("\\0");
else
printf("%c", a[i]);
#else
write(1, a, 6);
#endif
But you have to know about the 6.
The alternative is not to use NULL-terminated strings, and instead implement an alternative storage mechanism for your bytes; for example a length-encoded array.
#include <stdio.h>
typedef struct {
int length;
char *bytes;
} bytearr;
void my_printf(bytearr *arr)
{
#ifdef ESCAPE_NULLS
int i;
for (i = 0; i <= arr->length; i++)
if (arr->bytes[i] == 0)
printf("\\0");
else
printf("%c", arr->bytes[i]);
#else
write(1, arr->bytes, arr->length);
#endif
}
void main(void)
{
bytearr foo = {
6, "\0hey\0\0"
};
my_printf(&foo);
}
Graceless, but hopefully you get the idea.
Edit: 2011-05-31
Rereading the question I just noticed the word "concatenate". If the NULL characters are to be copied faithfully from one place in memory to another (not backslash-escape), and you know the total number of bytes in each array beforehand, then you can simply use memcpy.
#include <string.h>
char *a = "\0hey\0\0"; /* 6 */
char *b = "word\0up yo"; /* 10 */
char *c = "\0\0\0\0"; /* 4 */
void main(void)
{
char z[20];
char *zp = z;
zp = memcpy(zp, a, 6);
zp = memcpy(zp, b, 10);
zp = memcpy(zp, c, 4);
/* now z contains all 20 bytes, including 8 NULLs */
}

Answer 3

char *a="\0hey\0\0";
int alen = 7;
char buf[20] = {0};
int bufSize = 20;
int i=0;
int j=0;
while( i<bufSize && j<alen )
{
if(a[j]=='\0') {
buf[i++]='\\';
buf[i++]='0';
j++;
}
else {
buf[i++] = a[j++];
}
}
printf(buf);