In Bash, how do I test if a variable is defined in “-u” mode

Question

I just discovered set -u in bash and it helped me find several previously unseen bugs. But I also have a scenario where I need to test if a variable is defined

Answer 1

The answers above are not dynamic, e.g., how to test is variable with name "dummy" is defined? Try this:

is_var_defined()
{
    if [ $# -ne 1 ]
    then
        echo "Expected exactly one argument: variable name as string, e.g., 'my_var'"
        exit 1
    fi
    # Tricky.  Since Bash option 'set -u' may be enabled, we cannot directly test if a variable
    # is defined with this construct: [ ! -z "$var" ].  Instead, we must use default value
    # substitution with this construct: [ ! -z "${var:-}" ].  Normally, a default value follows the
    # operator ':-', but here we leave it blank for empty (null) string.  Finally, we need to
    # substitute the text from $1 as 'var'.  This is not allowed directly in Bash with this
    # construct: [ ! -z "${$1:-}" ].  We need to use indirection with eval operator.
    # Example: $1="var"
    # Expansion for eval operator: "[ ! -z \${$1:-} ]" -> "[ ! -z \${var:-} ]"
    # Code  execute: [ ! -z ${var:-} ]
    eval "[ ! -z \${$1:-} ]"
    return $?  # Pedantic.
}

Related: How to check if a variable is set in Bash?