Sending data to stdin of another process waiting on ppoll

Question

There is following code:

#include <poll.h>

int main(int argc, char *argv[])
{
  pollfd pd;
  pd.fd = 0;
  pd.events = POLLIN;
  int ret = ::ppoll(&pd, 1, 0, 0);
  return 0;
}

This should wait for some data on stdin and wake up when data appears.

Running program this way, it seems that ppoll is woken up (passing data to stdin via a pipe):

$ echo 1 | ./main
$

However sending something to the process from another shell doesn't wake up ppoll:

First shell:

$ ./main

Second shell:

$ echo 1 > /proc/$(pgrep main)/fd/0

After that I just see that in the first shell there is printed 1 , but ppoll doesn't wake up. How to send data correctly to a process's stdin from another shell so ppoll will wake up?

Answer 1

That's simply impossible, unless the process was specially launched in a way that enables feeding data to it from outside (ie through a FIFO or a pipe).

Even though /proc/$PID/fd/0 (ie standard input stream) points to the process's controlling terminal, so do (in the absence of redirection) /proc/$PID/fd/1 and /proc/$PID/fd/2 (ie standard output and standard error streams):

$ ls -l /proc/$$/fd
lrwx------ 1 user user 64 Sep  6 01:36 0 -> /dev/pts/3
lrwx------ 1 user user 64 Sep  6 01:36 1 -> /dev/pts/3
lrwx------ 1 user user 64 Sep  6 01:36 2 -> /dev/pts/3

The terminal is a pseudo-device, that can be both read from and written to. When a process reads from it, the terminal provides data entered from the keyboard or by other means (eg pasted from the clipboard). When you write to a terminal, it displays the data. Now when you write to /proc/$PID/fd/0 you write to the terminal rather than to the target process's standard input stream, that's why your data is displayed but the process stays indifferent.