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 variable expansion methods and programs like test and [ and [[, etc.

Right now I'm just looking for

DIR=$1 or . 

Meaning, my DIR variable should contain either what is specified in the first arg or the current directory.

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

I find the hyphen syntax confusing, and seek more readable syntax.

Why can't I do this?

DIR="$1" || '.' 

I'm guessing this means "if $1 is empty, the assignment still works (DIR becomes empty), so the invalid command '.' never gets executed."

Answer 1

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 '.'”.