Different output when printing a null char pointer with and without a newline appended

Question

I tried the following code on Linux

#include<stdio.h>
int main()
{
    char *p=NULL;
    printf("%s",p);
    return 0;
}

#include<stdio.h>
int main()
{
    char *p=NULL;
    printf("%s\n",p);
    return 0;
}

The first one outputs: (null)

While the second one causes a segmentation fault.

Why does \n make such a difference?

Hmm, I would have expected a `(null)` as well, but I also get the segfault. Replacing `\n` with something else, like `printf("%su", p)` prints `(null)u`. Even adding a space before `\n` prints `"(null) "`. Now I'm curious. — Pablo, Jan 14 '18 at 01:44
@Pablo It doesn't matter what result you get. The code is UB and may give any result. — Support Ukraine, Jan 14 '18 at 01:49
@Pablo Flag as duplicate with that link if you think it's more appropriate. — user4020527, Jan 14 '18 at 01:49
@4386427 I didn't that this was UB, using gcc I've always seen `(null)` being printed when passing `NULL` to `%s`, so I assumed that this was normal behaviour. Now I know better. — Pablo, Jan 14 '18 at 01:51
@Pablo Calling `printf(%s....", ptr);` with `ptr` being NULL is UB per standard — Support Ukraine, Jan 14 '18 at 01:52
@4386427 yes, I said that I didn't know that, hence my confusion. Now I know thanks to you and the link I found. — Pablo, Jan 14 '18 at 01:56

score 0 · Accepted Answer · answered Jan 14 '18 at 01:57

Both of your examples are undefined behavior per standard. Calling printf with %s and passing a NULL pointer is UB.

Therefore it makes no sense to discuss the outcome. On one system you might get one result and on another system you get another result.

Also see https://stackoverflow.com/a/11589479/4386427

Different output when printing a null char pointer with and without a newline appended

1 Answers1