Shell Programming and Scripting

Code:

abc$123

How do you see this? Do you see it as the letters abc plus the $ character and the numbers 123?
The shell see it as the string "abc', the positional variable deference $1 and the string "23"
Since there's nothing in $1 it assigns to par2 the string "abc23"

Code:

part=abc$xyz

The shell sees the string "abc" and the variable deference $xyz. And it appears that $xyz produces an empty string.

This is what Don said. Let us take apart the right side of the equation:

Code:

abc=> "abc" (literal string)
$1 => "" (because "$1" is not set, i.e. no argument is passed on commandline)
23 => "23" (literal string)

Therefore you get "abc23" with a "nothing" in between "abc" and "23".

Yes, because the difference is: "$xyz" is a legal variable name while "$123" is not. The variable "$xyz" might not be defined and therefore hold no content (which seems to be the case, because it evaluates to nothing), but it is still a legal variables name. "$123" is NOT a legal name, therefore the shell doesn't even try to evaluate it. Instead it uses what could be a legal name ("$1") and treats everything following it as a literal string again.

You could generate the same effect in your second example by surrounding the variables name with curly braces:

Code:

# v=abc${x}yz
# echo $v
abcyz

The same mechanism: "abc" is a literal string, the "$x" (which is empty, so ""), then "yz", which is again a literal string. If "x" is not empty:

Code:

# x=foo
# x=abc${x}yz
# echo $x
abcfooyz

I hope this helps.

bakunin

Thanks Don, Aia and Bakunin for your excellent explanation. I fully understand this now.


have a good day.