UNIX for Dummies Questions & Answers

Hi people I am trying to learn this code and see how it relates to the old DOS days. I have a line of code that I am not sure what the first part does. Any help will be greatly appreciated.

It is from a Save command that is used to backup files to a directory.
It goes like this

if [ $# -lt 1 ];then set /etc/ /etc/.save;fi

Thanks

If the argument count ($#) is less than (-lt) 1

Is is that what you meant?

That is sort of what I was looking for Jim but I am confused on the -lt part. I know this is a switch but not sure how it fits into the $#. And what does the 1 mean

$# is the argument list count

-lt means "less than"

1 means... well, one.

So, if the argument count given to the script is less than one (i.e. zero) then the
following block will be executed.

You should read up on what ==, -eq, -ne, -lt ... and so on mean in shell.

Scroll down to 'Conditional Expressions' here:

bash(1): GNU Bourne-Again SHell - Linux man page

Oh cool thanks Jim. I might be back with more questions

Great link Jim and I thank you but OK more questions LOL
So what everybody is saying is that if [$# -lt 1]; then set /etc/ /etc/.save; fi
if the $# is less than 1 then set will show the variables in the /etc/ file? right??? then how does the /etc/.save come into play?
The next line in the code is almost like the one I just posted except that it says
if [$# -lt 2] then set $* /etc/.save; fi
So if the argument count is less than 2 then it will show the variables in the $*????? I know the "*" is a metacharacter but there again how or what is it doing to the /etc/.save file? This stuff is making my head spin. It seems way over my head and I am supposed to learn it. There are just too many special characters that doing different things at different times and it is confusing the hell out of me. I thought C++ was bad but this takes the cake.