Bash-Scripting arrays

Question

I am new to bash-scripting & trying to understand how things work. It's all a bit strange..

The following can be put into a script or entered into the shell:

declare -a A=("foo" "bar")
B=1
[ ${A[B]} == ${A[$B]} ] && echo "wTF" || echo ";)"

This gives me "wTF" on my debian squeeze & also on cygwin 1.7.11-1

So. Why does ${A[B]} work?

@OliCharlesworth: I should have learned better than to argue with you after our last encounter! Still, I like bash. I like it a lot. — thb
– thb, Commented Mar 19, 2012 at 1:14
@thb: Each to their own! But IMO, any language that requires [ to be an executable in order to work is a language that sucks. — Oliver Charlesworth
– Oliver Charlesworth, Commented Mar 19, 2012 at 1:14
@OliCharlesworth: But [ isn't an executable, it's a builtin. (Granted, it's a deficient one -- the behavior of the [[ ... ]] notation is far superior -- but it's not as though changing your $PATH would suddenly break [.) — ruakh
– ruakh, Commented Mar 19, 2012 at 1:30
@violet313: [ and test are both builtins; neither is an alias for the other. They are equivalent in all respects, however, except that [ expects its last argument to be ] (which it discards). — ruakh
– ruakh, Commented Mar 19, 2012 at 1:40

ruakh · Accepted Answer · 2012-03-19 01:02:47Z

4

From the Bash Reference Manual, §6.7 "Arrays":

Indexed arrays are referenced using integers (including arithmetic expressions […]) and are zero-based; […] ¶ The subscript is treated as an arithmetic expression that must evaluate to a number greater than or equal to zero.

So in effect, ${A[B]} means ${A[$((B))]}. This is convenient when you want something like ${A[B-1]}.

Arithmetic expressions are explained in §6.5 "Shell Arithmetic", which says in part:

Within an expression, shell variables may also be referenced by name without using the parameter expansion syntax.

So, $((B)) means $(($B)) (except that the former is smarter about some things, e.g. using zero instead of blank as a default for uninitialized variables).

answered Mar 19, 2012 at 1:02

ruakh

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bkconrad · Accepted Answer · 2012-03-19 00:58:15Z

0

For the same reason A works: parameter expansion and substitution

answered Mar 19, 2012 at 0:58

bkconrad

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1 Comment

violet313 Over a year ago

Ok. I read that. I still don't get it. Suppose I use the following indirection: C=B; echo "${A[$C]}"; echo "${$C}" ?

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