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

@user657592 which bash version you are using ? and also check /usr/bin/timeout exists or not — Rahul Patil
– Rahul Patil, Commented Mar 28, 2014 at 14:08
Is that the only timeout you get if you run type -a timeout? In general you will need to export the function if you have it implemented somewhere to use it in a script, but like @RahulPatil mentioned timeout is usually just a c function unless you have some other implementations. — Reinstate Monica Please
– Reinstate Monica Please, Commented Mar 28, 2014 at 14:09
@user657592 Add export -f timeout to your .bashrc or wherever you have it implemented. And then re-source that init file (e.g. with source .bashrc) and try running the script again. — Reinstate Monica Please
– Reinstate Monica Please, Commented Mar 28, 2014 at 14:15

Community · Accepted Answer · 2017-04-13 12:36:31Z

5

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}"

Source

edited Apr 13, 2017 at 12:36

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answered Mar 28, 2014 at 13:59

Rahul Patil

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4 Comments

Rahul Patil Over a year ago

which bash version you are using ? and also check /usr/bin/timeout exists or not

user657592 Over a year ago

Using the absolute path does not help. It then says - /usr/bin/timeout: No such file or directory found

Rahul Patil Over a year ago

in your script, at the beginning

user657592 Over a year ago

Now, it says this - Starting AutoMATCH program alling Main-dispatch-ex program filename:

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