[SOLVED] Bash: how does this $[ ] code snippet work?

catkin · 09-09-2009, 07:53 AM

Hello

Why does bash do this

Code:

c@CW8:~$ a=b
c@CW8:~$ b=c
c@CW8:~$ c=d
c@CW8:~$ set -o nounset
c@CW8:~$ echo $[a]
bash: d: unbound variable
c@CW8:~$ set +o nounset
c@CW8:~$ echo $[a]
0
c@CW8:~$ echo $[a+1]
1

Best

Charles

RaptorX · 09-09-2009, 08:12 AM

hi catkin!

An unbound variable is one that has not been defined.

So lets review something...

Code:

[~]$ a=b
[~]$ echo $a
b
[~]$ echo ${a}
b
[~]$ echo $[a]
0

The only time i have seen [] in variables is in arrays like a[0] so maybe $[] is referencing a non existent array and is giving you the value 0.

If you are trying to reference to the "d" through "a" then i guess this would do the trick:

Code:

a=letter   # Variable "a" holds the name of another variable.
letter=z

echo

# Direct reference.
echo "a = $a"          # a = letter

# Indirect reference.
  eval a=\$$a
# ^^^        Forcing an eval(uation), and ...
#        ^   Escaping the first $ ...
# ------------------------------------------------------------------------
# The 'eval' forces an update of $a, sets it to the updated value of \$$a.
# So, we see why 'eval' so often shows up in indirect reference notation.
# ------------------------------------------------------------------------
  echo "Now a = $a"    # Now a = z

Now, remember that i am not an expert and this is just what i can see from the outside.

Im still learning.

catkin · 09-09-2009, 12:34 PM

Thanks RaptorX

Solved it! After intensive netsearching (not helped by Google ignoring string "$[" in searches!) the bash man page gave the answer:

Code:

   Arithmetic Expansion
       Arithmetic expansion allows the evaluation of an arithmetic expression and the substitution of the result.  The format  for  arithmetic  expansion
       is:

              $((expression))

       The old format $[expression] is deprecated and will be removed in upcoming versions of bash.

Best

Charles