Question

everybody! I have a simple question.
What does this line do?

trap "exec 1>&6 6>&- ; cat $LOGFILE" 0

So far, I understand that I am moving the standard output to 6 (this variable hasn’t been declared before, so I assume that 6 is a variable created in this same line), then the ‘6>&-‘ is something I don’t really get… and the cat $LOGFILE shows the contents of the variable LOGFILE. Also, the 0 at the end is supposed to mean that at the end of the execution of my program, execute ‘exec 1>&6 6>&- ; cat $LOGFILE’ before exiting.

Thanks for the help in advance!

Answer 1

1. trap <command> 0 means to execute <command> upon exit of the shell
2. exec 1>&6 means to redirect STDOUT (fd1) to fd6
3. exec 6>&- means to close fd6