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

How about this:

DIR=.
if [ $# -gt 0 ]; then
  DIR=$1
fi

$# is the number of arguments given to the script, and -gt means "greater than", so you basically set DIR to the default value, and if the user has specified an argument, then you set DIR to that instead.

Answer 2

I see several questions here.

1. “Can I write something that actually reflects this logic”

   Yes. There are a few ways you can do it. Here's one:

   if [[ "$1" != "" ]]; then
       DIR="$1"
   else
       DIR=.
   fi
   

2. “What is the difference between this and DIR=${1-.}?”

   The syntax ${1-.} expands to . if $1 is unset, but expands like $1 if $1 is set—even if $1 is set to the empty string.

   The syntax ${1:-.} expands to . if $1 is unset or is set to the empty string. It expands like $1 only if $1 is set to something other than the empty string.

3. “Why can't I do this? DIR="$1" || '.'”

   Because this is bash, not perl or ruby or some other language. (Pardon my snideness.)

   In bash, || separates entire commands (technically it separates pipelines). It doesn't separate expressions.

   So DIR="$1" || '.' means “execute DIR="$1", and if that exits with a non-zero exit code, execute '.'”.

Answer 3

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)