Why does using an unset variable as the array index return element 0 in Bash?

Question

Why is an empty key in a bash array always the zero key element?

Try

empty_key=

h[0]=0
h[1]=1
h[2]=2

echo ${h[$empty_key]}

Result

Could you please explain, why this behavior is correct?

In my understanding it should not be different to

not_existing_key=5 
echo ${h[$not_existing_key]}

where the result is just empty.

This is important to understand, if one uses the result for a loop, like in

for element in ${h[$key]};do
  ...
done

where the empty_key leads to a cycle, but NOT the not_existing_key.

To avoid this behavior it is obviously not to start an array with index 0

OR

replace an empty key by any key value not assigned (what seems to be boring).

Again my question: why is this behavior correct?

EDIT: My question should be understood as why is this behavior the prefered one in the bash world and not an empty result as with a non-existing key value?

@anubhava: Sorry, your statement is not true. A not_existing_key gives you just an empty result, not the 0-key element — onemorequestion
– onemorequestion, Commented Nov 28, 2019 at 15:59
Please, read again the question text where the not_existing_key has a non-empty value. Fue — onemorequestion
– onemorequestion, Commented Nov 28, 2019 at 17:57
Okay, but it does not make your first statement true and this should be clear to the next reader. Thx — onemorequestion
– onemorequestion, Commented Nov 28, 2019 at 18:10

Benjamin W. · Accepted Answer · 2019-11-28 15:40:42Z

7

When accessing an array element, [] constitutes an arithmetic context. This has a few consequences:

no need for $ to dereference variables:

$ arr=(a b c)
$ idx=1
$ echo ${arr[idx]}
b

arithmetic operations are evaluated:
```
$ echo ${arr[idx+idx]}
c
```
variable names are dereferenced in "chains" (evaluated as an arithmetic expression in the manual):
```
$ idx=2
$ idxref=idx
$ echo ${arr[idxref]}
c
```
names of null or unset variables are evaluated to 0 when used without parameter expansion syntax:
```
$ notset=
$ echo ${arr[notset]}
a
```
and finally (your question), a null value evaluates to 0:
```
$ notset=
$ echo {$arr[$notset]}
a
```

Manual quotes:

Arrays:

Indexed arrays are referenced using integers (including arithmetic expressions (see Shell Arithmetic)) [...]
Shell Arithmetic:

Shell variables are allowed as operands; parameter expansion is performed before the expression is evaluated. Within an expression, shell variables may also be referenced by name without using the parameter expansion syntax. A shell variable that is null or unset evaluates to 0 when referenced by name without using the parameter expansion syntax. [...] A null value evaluates to 0.

edited Nov 28, 2019 at 15:40

answered Nov 28, 2019 at 15:31

Benjamin W.

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11 Comments

KamilCuk Over a year ago

(your question), names of unset - hm... not really his question, he does ${h[$not_existing_key]} not ${h[not_existing_key]}. Or is it the same? I think this has to do with echo $[] printing 0. required for dereferencing - but doesn't the variable expansion inside the [ braces happens before variable expansion of ${ braces? So when ${ expansion happens it only sees empty [] braces, because $not_existing_key expands to nothing first?

Benjamin W. Over a year ago

@KamilCuk It's the same because of the first effect, no $ required for dereferencing.

Socowi Over a year ago

»variable names are dereferenced in "chains"«. It even goes beyond that: Whole expressions are evaluated recursively. Example: a=1; b=a+2; c=b+3; echo $((c)) will print the result of 1+2+3, that is, 6.

Benjamin W. Over a year ago

@KamilCuk I think "referencing without subscript" means $arr, which is the same as ${arr[0]}, though.

Benjamin W. Over a year ago

@onemorequestion "Why is this the way Bash does it" might only be answered by Chet Ramey.

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