Bash – Is checking for exit status other than 0 after a `command || return` unreachable

There is no nice way (I am aware of) to do that but if you are willing to pay the price...

Instead of putting the code in functions you can put it in files you source. If the functions need arguments then you have to prepare them with set:

set -- arg1 arg2 arg3
source ...

Three files:

1. testscript.sh

   #! /bin/bash
   
   startdir="$PWD"
   
   echo mainscript
   
   while true; do
           echo "begin: helper loop"
           source "${startdir}/func_1"
           echo "end: helper loop"
           break
   done
   
   echo mainscript
   

2. func_1

   echo "begin: func_1"
   source "${startdir}/func_2"
   echo "end: func_1"
   

3. func_2

   echo "begin: func_2"
   echo break from func_2
   break 100
   echo "end: func_2"
   

Result:

> ./testscript.sh
mainscript
begin: helper loop
begin: func_1
begin: func_2
break from func_2
mainscript

There's a few things here.

• You very seldom have to explicitly check $? against anything or save it in a variable (unless you need to reference the same exit status multiple times).
• The exit status of a function is the exit status of the last executed command in the function, so an explicit return is seldom needed (seldom with an explicit return value at least).
• A function that checks whether a directory exists should not create any directories. Better call it create_dir_if_needed.
• There's an error in [ result==0 ]. The string result==0 is a string of non-zero length, and testing a string in this way will return true if the string has non-zero length, so the test is always true. You probably wanted [ "$result" -eq 0 ] instead.
• Remember to always double quote variable expansions and command substitutions, unless you know in what contexts this is not needed.

With these things in mind:

create_dir_if_needed () {
    mkdir -p -- "$1"
}

This would return the exit status of mkdir -p -- "$1". This command would create the named directory (and any intermediate directories) if this did not already exist. If the mkdir command fails to create the directory, it will exit with a non-zero exit status, which will become the exit status of the function. mkdir -p will not fail if the directory already exists.

You would use this as

if ! create_dir_if_needed "$dirpath"; then
    printf 'Failed to create directory "%s"\n' "$dirpath" >&2
    exit 1
fi

or, since the function is trivial, you could get rid of it and say

if ! mkdir -p -- "$dirpath"; then
    printf 'Failed to create directory "%s"\n' "$dirpath" >&2
    exit 1
fi

A variation of the create_dir_if_needed function that uses mkdir without -p and will therefore never create missing parent directories to the given directory path:

create_dir_if_needed () {
    if [ -d "$1" ]; then
        return
    fi

    mkdir -- "$1"
}

or,

create_dir_if_needed () {
    [ -d "$1" ] || mkdir -- "$1"
}

A call to this function would return true (zero) if the directory already existed or if the mkdir call went well. A return statement with no explicit value will return the exit status of the most recently executed statement, in this case it would return the positive outcome of the [ -d "$1" ] test.