Stopping script from nested function [duplicate]

Question

I have this script:

#!/bin/bash

function_1() {
    function_2
    printf world
}

function_2() {
    printf "hello "
    return 1
    printf "not printed"
}

function_1

which prints hello world. I would like to stop the execution of the script in function_2 after the first printf. So the script should only print hello. I tried return 1 but this only returns back to function_1. I also tried to send a kill signal as proposed in earlier answer but I got the same result as with return 1.

nullPointer · Accepted Answer · 2019-04-17 06:45:45Z

0

you can add an exit 0 to exit the script with a 0 exit status at the desired point :

function_2() {
    printf "hello "
    exit 0
    ...

answered Apr 17, 2019 at 6:45

nullPointer

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1 Comment

builder-7000 Over a year ago

Thank that works. But if I put these two functions in .bashrc then call function_1 from the terminal, the terminal gets closed.

Daniel H · Accepted Answer · 2019-04-17 06:45:24Z

0

I believe you want the exit built-in. From the Bash manpage:

exit [n]

Cause the shell to exit with a status of n. If n is omitted, the exit status is that of the last command executed. A trap on EXIT is executed before the shell terminates.

answered Apr 17, 2019 at 6:45

Daniel H

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