Bash: How to set a variable from argument, and with a default value

Question

It is pretty clear that with shell scripting this sort of thing can be accomplished in a huge number of ways (more than most programming languages) because of all the different

Answer 1

I use a simple helper function to make such assignments look cleaner. The function below accepts any number of arguments, but returns the first one that's not the empty string.

default_value() {
    # Return the first non-empty argument
    while [[ "$1" == "" ]] && [[ "$#" -gt "0" ]]; do
        shift
    done
    echo $1
}
x=$(default_value "$1" 0)