set variable in subshell

Question

I have a script like this that works fine:

#!/bin/bash

INBOUND=$(echo '5')
OUTBOUND=$(echo '10')

TOTAL=$(($INBOUND+$OUTBOUND))

echo "IN:$INBOUND OUT:$OUTBOUND T:$TOTAL"

Output: IN:5 OUT:10 T:15

Now suppose instead of echo 5 and echo 10 I have two commands that takes 10 seconds each to run.

I don't want my script to take 10+10 seconds to run and so I'm trying to use two sub-processes for each variable.

I thought whis would work:

#!/bin/bash

INBOUND=$(echo '5') &
OUTBOUND=$(echo '10')
wait
TOTAL=$(($INBOUND+$OUTBOUND))

echo "IN:$INBOUND OUT:$OUTBOUND T:$TOTAL"

But the first variable doesn't get a value and the output is: IN: OUT:10 T:10

How can I set each var in a separate process, so the script runs in 10 seconds instead of 10+10 ?

http://stackoverflow.com/questions/20017805/bash-capture-output-of-command-run-in-background — dekkard, Mar 22 '16 at 17:17

score 3 · Accepted Answer · answered Mar 22 '16 at 16:30

3

You need to use (temporary) files.

inbound_result=$(mktemp)
outbound_result=$(mktemp)
echo '5' > "$inbound_result" &
echo '10' > "$outbound_result" &
wait
read inbound < "$inbound_result"
read outbound < "$outbound_result"
total=$((inbound + outbound))

echo "IN: $inbound OUT:$outbound T:$total"

answered Mar 22 '16 at 16:30

chepner

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Worked like a charm. Thank you! – Svperstar Mar 22 '16 at 17:11

set variable in subshell

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