why printf works on non-terminated string

Question

I am wondering how does printf() figure out when to stop printing a string, even I haven\'t put a termination character at the end of the string? I did an experiment by mall

Answer 1

Is because you had bad lucky and the next byte of your malloc'ed string was a 0 byte.

You can confirm that by doing:

const char* digits = "0123456789";
char* buff = (char*)malloc(10);
memcpy(buff, digits, 10);

printf("%s, %d\n", buff, (int)*(buff + 10));

Your program have to print:

0123456789 0

And that 0 is the NULL which you did't malloc'ed but it was there. Note that this behavior is UNDEFINED so you cannot trust in these things. As I said before this happened because you are unlucky! The good thing to happen in this situation is a SIGSEGV.