Assign array to variable

Question

The issue about assigning arrays to variables in bash script seems rather complicated:

a=(\"a\" \"b\" \"c\")
b=$a

echo ${a[0]} 
echo ${a[1]}

echo ${b[0]} 
echo         


        

Answer 1

Why?

If a is an array, $a expands to the first element in the array. That is why b in your example only has one value. In bash, variables that refer to arrays aren't assignable like pointers would be in C++ or Java. Instead variables expand (as in Parameter Expansion) into strings and those strings are copied and associated with the variable being assigned.

How can I fix it?

To copy a sparse array that contains values with spaces, the array must be copied one element at a time by the indices - which can be obtained with ${!a[@]}.

declare -a b=()
for i in ${!a[@]}; do
    b[$i]="${a[$i]}"
done

From the bash man page:

It is possible to obtain the keys (indices) of an array as well as the values. ${!name[@]} and ${!name[*]} expand to the indices assigned in array variable name. The treatment when in double quotes is similar to the expansion of the special parameters @ and * within double quotes.

Here's a script you can test on your own:

#!/bin/bash

declare -a a=();
a[1]='red hat'
a[3]='fedora core'

declare -a b=();

# Copy method that works for sparse arrays with spaces in the values.
for i in ${!a[@]}; do
    b[$i]="${a[$i]}"
done

# does not work, but as LeVar Burton says ...
#b=("${a[@]}")

echo a indicies: ${!a[@]}
echo b indicies: ${!b[@]}

echo "values in b:"
for u in "${b[@]}"; do
    echo $u
done

Prints:

a indicies: 1 3
b indicies: 1 3  # or 0 1 with line uncommented
values in b:
red hat
fedora core

This also works for associative arrays in bash 4, if you use declare -A (with capital A instead of lower case) when declaring the arrays.

Answer 2

If you want to copy a variable that holds an array to another name, you do it like this:

a=('a' 'b' 'c')
b=( "${a[@]}" )