The break builtin for bash does allow you to accomplish what you are doing, just break with a negative value and the status returned by $? will be 1:
while true
do
if [ -f "./test" ] ; then
break -1
fi
done
echo $? ## You'll get 1 here..
Note, this is documented in the help for the break builtin:
help break
break: break [n] Exit for, while, or until loops.
Exit a FOR, WHILE or UNTIL loop. If N is specified, break N enclosing
loops.
Exit Status: The exit status is 0 unless N is not greater than or
equal to 1.
You can break out of n number of loops or send a negative value for breaking with a non zero return, ie, 1
I agree with @hagello as one option doing a sleep and changing the loop:
#!/bin/bash
timeout=120
waittime=0
sleepinterval=3
until [[ -f "./test" || ($waittime -eq $timeout) ]]
do
$(sleep $sleepinterval)
waittime=$((waittime + sleepinterval))
echo "waittime is $waittime"
done
if [ $waittime -lt $sleepinterval ]; then
echo "file already exists"
elif [ $waittime -lt $timeout ]; then
echo "waited between $((waittime-3)) and $waittime seconds for this to finish..."
else
echo "operation timed out..."
fi