Re: [Help-bash] inconsistency in expansions

[John Kearney]

Am 07.05.2012 15:12, schrieb Greg Wooledge:

On Mon, May 07, 2012 at 07:09:15AM -0600, Bill Gradwohl wrote:

${$x} isn't a valid expansion, so it throws an error, the pattern isn't
matched, and nothing happens.

I see it produces an error, but I still don't see why. If $x expands to 3
why don't I get ${3} and then the expansion of ${3}? Why is it illegal?
What bash rule makes it so?

The definition of parameter expansion:

$parameter
${parameter[operator]}

$x is not a valid parameter.  x is a valid parameter.  3 is a valid
parameter.  $x is not.  Therefore you cannot write ${$x}.

You just got me to realize that these work
both
${!x}
and
${!#}

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