[SOLVED] How can I quote the last argument in the bash script?

[phoenic]

I've tried made it like this:
#! /bin/bash

echo $[$#]

But it seems the outer $ doesn't work, I want the result like this:

./test a f d s g

g

How can I make it?

[drl]

Hi.

I recommend that you look at: http://www.tldp.org/LDP/abs/html/othertypesv.html around the text:

Bracket notation for positional parameters leads to a fairly simple way of referencing the last argument passed to a script on the command line ...

In fact, the guide http://www.tldp.org/LDP/abs/html/index.html is something that you can use to find almost every answer for bash questions ... cheers, drl

[phoenic]

Ok, I've got it.
just add "eval" at the head.

eval echo ${$#}