Which stream does “stack smashing detected” message get printed to?

Question

Consider the following very basic program, which has appeared in many forms on other questions here.

#include <string.h>

int main() {
    char message[8];
    strcpy(message, "Hello, world!");
}

On my system, if I put this in a file called Classic.c, compile it with no special flags and run it, I get the following output.

$ gcc -o Classic Class.c 
$ ./Classic
*** stack smashing detected ***: ./Classic terminated
Aborted (core dumped)

Normally, program output goes to stderr or stdout, so I expected that the following would produce no output.

./Classic  2> /dev/null > /dev/null

However, the output is exactly the same, so I have three questions to this scenario.

1. What stream is being printed to here?
2. How could I write code that prints to this special stream (without smashing my stack deliberately).
3. How can I redirect the output of this stream?

Note I am running on a Linux system. Specifically, Ubuntu 14.04.

Answer 1

Since it's not stderr or stdout, there's only one remaining option: The controlling tty.

You can write to this with your code by opening /dev/tty.

Redirecting its output is intentionally very difficult (this is why /dev/tty is also used for password prompts). That said, if you really want to do it, expect can be used for this purpose, as can emPTY.

---

The easiest approach with expect is to use the included helper unbuffer, which will effectively redirect this content to stdout:

$ sh -c 'echo hello >/dev/tty' >/dev/null 2>&1
hello
$ unbuffer sh -c 'echo hello >/dev/tty' >/dev/null 2>&1
$