#!/bin/bash
# query1.sh
numbers=(53 8 12 9 784 69)
echo ${numbers[3]} # <--- this echoes "9" to standard output.
number=numbers[3]
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?
Your problem is that this:
number=numbers[3]
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]
53[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
9
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}"
9
The zsh equivalent would be:
numbers=(53 8 12 9 784 69)
number=numbers[4]
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).
9 directly, whereas this is more like a workaround for their incorrect guess.
echo ${numbers[3]} in the question).
${numbers[3]} to another variable.
$i, where i=3)? The syntax has to change again..
numbers[i].
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]}
