Why does timeout not work within a bash script? - linux

Question

I tried to kill a process if it exceeds more than a few seconds.
The following works just fine when I run it in the terminal.
timeout 2 sleep 5
But when I have a script -
#!/bin/bash
timeout 2 sleep 5
it says
timeout: command not found
Why so? What is the workaround?
--EDIT--
On executing type timeout, it says -
timeout is a shell function

Answer 1

It's seems your environment $PATH variable does not include /usr/bin/ path or may be timeout binary exists in somewhere else.
So just check path of timeout command using :
command -v timeout
and use absolute path in your script
Ex.
#!/bin/bash
/usr/bin/timeout 2 sleep 5
Update 1#
As per your update in question, it is function created in your shell. you can use absolute path in your script as mentioned in above example.
Update 2#
timeout command added from coreutils version => 8.12.197-032bb, If GNU timeout is not available you can use expect (Mac OS X, BSD, ... do not usually have GNU tools and utilities by default).
################################################################################
# Executes command with a timeout
# Params:
# $1 timeout in seconds
# $2 command
# Returns 1 if timed out 0 otherwise
timeout() {
time=$1
# start the command in a subshell to avoid problem with pipes
# (spawn accepts one command)
command="/bin/sh -c \"$2\""
expect -c "set echo \"-noecho\"; set timeout $time; spawn -noecho $command; expect timeout { exit 1 } eof { exit 0 }"
if [ $? = 1 ] ; then
echo "Timeout after ${time} seconds"
fi
}
Example:
timeout 10 "ls ${HOME}"
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