Why does timeout not work within a bash script?

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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