# query1.sh

numbers=(53 8 12 9 784 69)
echo ${numbers[3]} # <--- this echoes "9" to standard output.

echo number # <--- this echoes "number" to stdout.
echo $number # <--- this echoes "numbers[3]" to stdout.
echo ... <--- ???

What syntax should I use to echo the variable named number and get "9" as the standard output?


That's where you want variable indirection using the ${!var} operator:

bash-4.4$ numbers=(53 8 12 9 784 69)
bash-4.4$ number=numbers[3]
bash-4.4$ echo "${!number}"

The zsh equivalent would be:

numbers=(53 8 12 9 784 69)
echo ${(P)number}

(zsh arrays indices start at 1 like in most other shells and shell tools, so you can use numbers[4] to get the 4th number).

If you wanted to assign the value of 4th element of the array to $number, you'd do number=$numbers[4] in zsh, or number=${numbers[3]} in ksh (and bash which just copied ksh arrays).

  • Sure, it can be done by using indirection, but I wouldn't go as far as to say that it's what the OP wants. It seems more reasonable to expect that what they want is the correct syntax to get the element 9 directly, whereas this is more like a workaround for their incorrect guess. – Tom Fenech Sep 18 '17 at 19:10
  • @Tom, that would be the last part of the answer. But it looks like they known already how to get the value of an element of the array (echo ${numbers[3]} in the question). – Stéphane Chazelas Sep 18 '17 at 19:27
  • Yeah, good point. My understanding was that they hadn't worked out that they could assign ${numbers[3]} to another variable. – Tom Fenech Sep 18 '17 at 19:31
  • @Stéphane: Many thanks. Further to this: What if the array's index is a variable (let's say $i, where i=3)? The syntax has to change again.. – Anthony Webber Sep 18 '17 at 20:12
  • @AnthonyWebber, the array index is taken as an arithmetic expression, so you can use numbers[i]. – Stéphane Chazelas Sep 18 '17 at 20:17

Your problem is that this:


Is not assigning the 4th element of the array numbers to the variable number. That's because numbers[3] doesn't actually mean anything:

$ numbers=(53 8 12 9 784 69)
$ echo $numbers[3]

What happens is that when you use the name of the array alone as a variable, what is actually being used is the first element of the array (53). So $numbers[3] becomes 53[3]. What you want is:

$ number=${numbers[3]}
$ echo $number

When you declare a variable, you don't need to use the $, but when you use it, you should.

So just assign like this, and it should works number=${numbers[3]}

  • That won't and doesn't work, and isn't what I asked. But I appreciate you answering. – Anthony Webber Sep 18 '17 at 14:08
  • Yep, i made a mistake in the syntax. I corrected it. – Carpette Sep 18 '17 at 14:54

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