If $# is a number, aren’t "$#" and $# exactly the same in Bash? Enclosing things in double quotes doesn’t change their type, since everything is already treated as a string.

https://www.gnu.org/software/bash/manual/html_node/Double-Qu...

While the other reply to you is correct, the actual answer to your question is No -- bash interprets leading 0s (zeros) in the C sense, it indicates octal not decimal. So this very small bash will fail for you:

    X="08"
    if [[ "$X" -eq 8 ]]; then
      echo Y
    fi

Use `bash -x` to run it in debug mode:

    $ bash -x test.sh 
    + X=08
    + [[ 08 -eq 8 ]]
    test.sh: line 2: [[: 08: value too great for base (error token is "08")

This comes into play if you're dealing with things like IPv4 address, where it is technically allowable to have leading zeros, however by convention it's never done because of code traps which see those leading zeros as octal mode, not decimal.

Your example only demonstrates the effect of `[[`. If you replace it with `[` the error won't occur. In both cases double-quoting (or not) `$X` makes no difference in the context of your example.

[ is an external program called "test" and [[ is a bash builtin, they have many different behaviours beyond just trying to use one to prove the other wrong on a random internet comment.

I merely used that to show that the error message was tied to `[[` and not to the use of double quotes (which your parent comment was asking about).

It's true, but you're semantically overloading the variable syntax (by hinting that it's treated as string).

Plus, left and single operands, in bash conditions, don't require quoting, so a lean style is actually not to use them at all.

> …left and single operands…

That's false:

  x='foo bar'; [ $x = 'foo bar' ]; echo $?
  -bash: [: too many arguments

  x='foo bar'; [ "$x" = 'foo bar' ]; echo $?
  0

As I replied to your other comment, you cannot just replace [[ with [ and get the same behaviour. If you use the actual bash builtin [[ which we're discussing, it works as expected. Also, please use == when comparing, = is an assignment operator (in almost every programming language).

My bad. I interpreted “bash conditions” differently.

Is this actually a shellcheck error? It just evaluates to the integer number of arguments in $@, so I doubt it is (and I doubt it matters whether it's quoted? unless there's a bash back-compat issue here?)

Is there a bug in the command line parsing there? I can't see it.

In this specific case it's more of a nitpick, I think. Even shellcheck won't demand double-quoting `$#` because everyone(?) knows that it'll always be a number and have no whitespace in it nor evaluate to an empty string.