Shell Programming and Scripting

Hi,
I picked up a shell script function to implement a read timeout functionality : Wait for specified number of seconds to read input, else return.

E.g:
read_timeout 5 localvar
This would wait for 5 seconds to get a user input, and will exit if no input is given for 5 seconds. The localvar variable would become null in the timed out case.

The code goes as :

read_timeout() {
trap : USR1
trap 'kill "$pid" 2> /dev/null' EXIT
(sleep "$1" && kill -USR1 "$$") & pid=$!
read "$2"
ret=$?
kill "$pid" 2> /dev/null
trap - EXIT
return "$ret"
}

Inside this function, I want to set the input variable to a default value if the time out happens.

E.g. :
With the statement
read_timeout 5 localvar

localvar should become equal to "nothing" if I do not enter any value for 5 seconds. This is not happening if a set localvar = nothing before invoking the read_timeout function

Is it possible with this piece of code?


Thanks,
Puneet Arora

bash has a read -t seconds builtin. It does exacatly what you want.