Is this a correct way to test if a parameter is declared?

Question

Here is an example from https://unix.stackexchange.com/a/56846/674

A different, bash-specific way of testing whether a variable of any type has been declared is to check whether it's listed in ${!PREFIX*}:

case " ${!foobar*} " in
  *" foobar "*) echo "foobar is declared";;
  *) echo "foobar is not declared";;
esac

1. If I am correct, to make a parameter not declared, we use unset builtin. The unset case works like the quote said,

   $ unset foobar
   $ case " ${!foobar*} " in
   >   *" foobar "*) echo "foobar is declared";;
   >   *) echo "foobar is not declared";;
   > esac
   foobar is not declared
   

   but the declared case doesn't:

   $ unset foobar
   $ declare foobar
   $ case " ${!foobar*} " in
   >   *" foobar "*) echo "foobar is declared";;
   >   *) echo "foobar is not declared";;
   > esac
   foobar is not declared
   

   Is this a correct way to test if a parameter is declared?

2. Also in " ${!foobar*} ", why are there spaces before and after the parameter expansion? Does it work when foobar is an array or dictinary and when it is a variable but not an array or dictionary?

Answer 1

1. Looks like you'd have to declare the variable with a value (can be null):

   unset foobar; declare foobar=
   

   The case statement will produce "foobar is declared".

2. you use quotes and spaces so the *" foobar "* can detect the specific variable name out of the list of varnames returned by ${!foobar*}:

   $ unset foobar; foobarbaz=1; foobarquz=2
   $ echo " ${!foobar*} "
    foobarbaz foobarquz 
   

   The case statement will produce "foobar is not declared".