How to get the exit status a loop in bash

Question

I know how to check the status of the previously executed command using $?, and we can make that status using exit command. But for the loops in bash are always returning a

Answer 1

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

Answer 2

The bash manual says:

while list-1; do list-2; done
until list-1; do list-2; done
  [..]The exit status of the while and until commands is the exit status
  of the last command executed in list-2, or zero if none was executed.[..]

The last command that is executed inside the loop is break. And the exit value of break is 0 (see: help break).

This is why your program keeps exiting with 0.