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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?
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Tim
  • 101,790

1 Answers1

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  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".

glenn jackman
  • 85,964
  • Thanks. I am a little confused. What does *" foobar "* expand to? Does * match any string, so foobarbaz should be in *" foobar "*? – Tim May 02 '16 at 22:02
  • In that context, it's a pattern where * will match zero or more of any character. We are checking if the string " foobar " is contained in the string " ${!foobar*} " – glenn jackman May 02 '16 at 22:12