Re: [Help-bash] Arithmetic evaluation / expansion question

[Greg Wooledge]

On Thu, Oct 11, 2018 at 10:33:55AM -0400, Chet Ramey wrote:
> No, Greg. If the evaluator encounters a token that looks like a shell
> identifier, its value is treated as an expression (if it's unset, that
> expression evaluates to 0). This expression is run through the evaluator
> again, as if it had been seen inline in parens. This is what makes
> 
> ten='2 * 5'
> echo $(( ten ))
> echo $(( $ten ))
> 
> both echo `10'. I'm surprised you said that, because this has come up a
> number of times in the past, but your wiki has the same error.

OK, thanks.  Corrected it on the wiki page.
https://mywiki.wooledge.org/ArithmeticExpression