How to modify a global variable within a function in bash?

Question

I\'m working with this:

GNU bash, version 4.1.2(1)-release (x86_64-redhat-linux-gnu)

I have a script like below:

#!/bin/bas         


        

Answer 1

When you use a command substitution (ie the $(...) construct), you are creating a subshell. Subshells inherit variables from their parent shells, but this only works one way - a subshell cannot modify the environment of its parent shell. Your variable e is set within a subshell, but not the parent shell. There are two ways to pass values from a subshell to its parent. First, you can output something to stdout, then capture it with a command substitution:

myfunc() {
    echo "Hello"
}

var="$(myfunc)"

echo "$var"

Gives:

Hello

For a numerical value from 0-255, you can use return to pass the number as the exit status:

mysecondfunc() {
    echo "Hello"
    return 4
}

var="$(mysecondfunc)"
num_var=$?

echo "$var - num is $num_var"

Gives:

Hello - num is 4