How to copy an array in Bash?

Question

I have an array of applications, initialized like this:

depends=$(cat ~/Depends.txt)

When I try to parse the list and copy it to a new arra

Answer 1

a=(foo bar "foo 1" "bar two")  #create an array
b=("${a[@]}")                  #copy the array in another one 

for value in "${b[@]}" ; do    #print the new array 
echo "$value" 
done   

Answer 2

The simplest way to copy a non-associative array in bash is to:

arrayClone=("${oldArray[@]}")

or to add elements to a preexistent array:

someArray+=("${oldArray[@]}")

Newlines/spaces/IFS in the elements will be preserved.

For copying associative arrays, Isaac's solutions work great.

Answer 3

The solutions given in the other answers won't work for associative arrays, or for arrays with non-contiguous indices. Here are is a more general solution:

declare -A arr=([this]=hello [\'that\']=world [theother]='and "goodbye"!')
temp=$(declare -p arr)
eval "${temp/arr=/newarr=}"

diff <(echo "$temp") <(declare -p newarr | sed 's/newarr=/arr=/')
# no output

And another:

declare -A arr=([this]=hello [\'that\']=world [theother]='and "goodbye"!')
declare -A newarr
for idx in "${!arr[@]}"; do
    newarr[$idx]=${arr[$idx]}
done

diff <(echo "$temp") <(declare -p newarr | sed 's/newarr=/arr=/')
# no output

Answer 4

Try this: arrayClone=("${oldArray[@]}")

This works easily.

Answer 5

Starting with Bash 4.3, you can do this

$ alpha=(bravo charlie 'delta  3' '' foxtrot)

$ declare -n golf=alpha

$ echo "${golf[2]}"
delta  3

Answer 6

You can copy an array by inserting the elements of the first array into the copy by specifying the index:

#!/bin/bash

array=( One Two Three Go! );
array_copy( );

let j=0;
for (( i=0; i<${#array[@]}; i++)
do
    if [[ $i -ne 1 ]]; then # change the test here to your 'isn't installed' test
        array_copy[$j]="${array[$i]}
        let i+=1;
    fi
done

for k in "${array_copy[@]}"; do
    echo $k
done

The output of this would be:

One
Three
Go!

A useful document on bash arrays is on TLDP.