bash: function returning output instead of output status

Question

I have a small bash script as below

function getUptime() {

    uptime 2>/dev/null
    return "$?"
}

resultReceived=$(getUptime)
echo "Result: $resultReceived"

And when I execute the script, instead of 0 I am getting uptime command output. Where I am doing the mistake. Please help.

Debug Output

 tmp bash -x testingscript.sh
++ getUptime
++ uptime
++ return 0
+ resultReceived='11:47  up 3 days, 19:24, 7 users, load averages: 1.88 1.78 2.04'
+ echo 'Result: 11:47  up 3 days, 19:24, 7 users, load averages: 1.88 1.78 2.04'
Result: 11:47  up 3 days, 19:24, 7 users, load averages: 1.88 1.78 2.04

https://stackoverflow.com/questions/17336915/return-value-in-a-bash-function — Mat, Dec 04 '19 at 06:37

jeb · Accepted Answer · 2019-12-04T06:42:33.510

You mixed the exit status and the output.
With $(getUptime) you get the output of your function, not the exit status.

You could change your function, but then the name will not match it's functionality.

function getUptime() {

    uptime >/dev/null 2>/dev/null
    echo "$?"
}

I suppose it's better not to change the function, instead change the code at:

resultReceived=$(getUptime)
exitstatus=$?
echo "Result: $resultReceived, exitstatus: $exitstatus"

bash: function returning output instead of output status

1 Answers1