Copying a Bash array fails

Question

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 ${b[1]}  

leads to

a  b  a     

instead of

a b a b 

Why? How can I fix it?

Answer 1

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[@]}" ) 

Answer 2

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.