Thank you very much. Actually, I finally found it for myself not long after I posted. There's a good example of it on the "special variables" page, referencing positional variables. But by the time I came back, I had two answers. You guys are fast.

I see now that it's an interesting use of indirect referencing...it outputs the value of the last positional variable, as given by "$#". Easy once you know what the parts mean.

But now why is it when I try it with the old syntax, and use 'echo \$$#', it doesn't work? It only give me the output "$3" (for example) when I try it, instead of the actual value.

Edit: Found the reason, I think. You need to eval the older notation before it will work. 'eval echo \$$#' seems to do the trick. I still don't quite understand exactly why you need to do this though.

You have to use the old syntax for indirect reference in conjunction to the eval built-in. Running
the shell substitutes the variable $# with the actual number of arguments and interprets the escaped $ sign as a litteral $. Using eval instead
first the shell does the above substitution, then passes the resulting string "echo $3" to the eval bult-in, which in turn causes the substitution of $3 with its actual value.

Edit: you are faster than us, now!

Thank you for the explanation. I kind of suspected that that was what was happening, but I wasn't sure. I'm still not quite clear on exactly what things like eval do sometimes.

I suppose the new ${!} notation bypasses all that and just grabs the value directly then? It's certainly easier and much more intuitive in any case.

Deleted. Accidentally duplicated my post.