Removal of the nohup -file during the execution

Question

I did the following:

nohup find / &

rm nohup.out

Oddly, the nohup -command continued to run. I awaited for a new file to be created. For my surprise there was no such file. Where did the stdout of the command go?

Answer 1

Removing a file in UNIX does two things:

• it removes the directory entry for it.
• if no processes have it open and no other directory entries point to it (hard links), it releases the space.

Your nohupped process will gladly continue to write to the file that used to be called nohup.out, but is now known as nothing but a file descriptor within that process.

You can even have another process create a nohup.out, it won't interfere with the first.

When all hard links are gone, and all processes have closed it, the disk space will be recovered.

Answer 2

if you will delete the nohup.out file, the handle will be lost and it will write only to the file descriptor but if you want to clean out the nohup.out file then just run this

true > nohup.out

it will delete the contents of the file but not the file.

Answer 3

That's standard behaviour on Unix.

When you removed the file, the last link in the file system was removed, but the file was still open and therefore the output of find (in this case) was written to disk blocks in the kernel buffer pool, and possibly even to disk. But the file had no name. When find exited, no process or file (inode) referenced the file, so the space was released. This is one way that temporary files that will vanish when a program exits are created - by opening the file and then removing it. (This presumes you do not need a name for the file; clearly, if you need a name for the temporary, this technique won't work.)

Answer 4

cat /dev/null > nohup.out

from here