如何在退出时删除/取消链接文件

Question

How do I make, in Node, so that when the process is terminated or when the execution is halted a certain file I open with fs.openSync(filename, 'w'); is deleted/unlinked?

On windows, If I create the file and then I call fs.unlinkSync(filename); right after, the other program can see the file. Additionally, when the program that created it terminates, the file is also gone.

On Linux, the above is not true because after I unlink it, when the other tries to check the existence of the file it cannot find it in any shape or form I know of.
Locks are not enforced unless the drive is mounted in a specific way.
If I try to find a way to call fs.unlinkSync(filename); in callbacks either to SIGINT , SIGTSTP , etc... I always find a way to stop the program without the file being gone afterwards.

Please help me finding a way to create such file where being there is directly related to those files being in use.

(please ignore good practices and conventions, for now).

Answer 1

Consider a shell script like:

#!/bin/sh

trap '' INT HUP TERM

read fname
read dummy

[ -f "$fname" ] && rm -f "$fname"

If you use popen(3) (or Node's equivalent) to open a pipe to that script, write a pathname on the pipe, and then do nothing else to that pipe , when your application exits, the invoked script will delete the file, having seen EOF on the "read dummy" line, because your app's exit will have closed the pipe to the invoked script.

I've seen similar approaches used from Java apps that really want to see some temporary directory hierarchies cleaned up on JVM exit: have the helper script do it. Just make sure that the pipe to the helper is not closed while your main application runs.

This likely fits your "ignore good practices and conventions" suggestion...