问题
This works fine (infinite loop):
$ while TRUE; do printf ".";done
.............................................................................
I am trying to timeout this while loop with the timeout command.
All these don't work:
$ timeout 5 while TRUE; do printf ".";done
$ timeout 5 "while TRUE; do printf ".";done"
$ timeout 5 "while TRUE; do printf \".\";done"
$ timeout 5 $(while TRUE; do printf ".";done)
$ timeout 5 $('while TRUE; do printf ".";done')
What is the correct way (if it exists)?
回答1:
I think that the solution to your problem is to execute another shell instance and pass proper commands to it. According to bash manual:
-c If the -c option is present, then commands are read from the first non-option argument command_string.
Thus my solution would be something like that:
timeout 5 bash -c -- 'while true; do printf ".";done'
-- assures that the following arguments will be treated as non-option. And '' helps with passing " without unnecessary escaping
来源：https://stackoverflow.com/questions/27555727/timeouting-a-while-loop-in-linux-shell-script