Bash script print positional parameters

[belze]

Im taking a class on Unix/Linux and now we have reached a point in the class where I have no experience: Scripting.
Im trying to make a bash script for a homework excercise. I need to make a script that prints out 10 positional parameters. After some research I found out that by default there are 10, but 0 is actually the name of the script so I believe that wont count for this excercise.
I looked a little more and found the shift command.

This is what I have so far:
#!/bin/bash

echo $1
echo $2
echo $3
echo $4
echo $5
echo $6
echo $7
echo $8
echo $9
shift 1
echo $9

This gets the job done, but is this the correct/easiest way to do what Im trying to do? I would like to get them all on one line but with the shift command in there I havent been able to so far.

[Franklin52]

Use a for loop with the seq command and shift the parameters after every "echo $1".

Have a read of these guides:

http://tldp.org/LDP/Bash-Beginners-G...tml/index.html
http://www.tldp.org/LDP/abs/html/index.html

and check the manpage of seq.

Regards