How to copy an array in Bash? - arrays

Question

I have an array of applications, initialized like this:
depends=$(cat ~/Depends.txt)
When I try to parse the list and copy it to a new array using,
for i in "${depends[#]}"; do
if [ $i #isn't installed ]; then
newDepends+=("$i")
fi
done
What happens is that only the first element of depends winds up on newDepends.
for i in "${newDepends[#]}"; do
echo $i
done
^^ This would output just one thing. So I'm trying to figure out why my for loop is is only moving the first element. The whole list is originally on depends, so it's not that, but I'm all out of ideas.

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

array_copy() {
set -- "$(declare -p $1)" "$2"
eval "$2=${1#*=}"
}
# Usage examples:
these=(apple banana catalog dormant eagle fruit goose hat icicle)
array_copy these those
declare -p those
declare -A src dest
source=(["It's a 15\" spike"]="and it's 1\" thick" [foo]=bar [baz]=qux)
array_copy src dest
declare -p dest
Note: when copying associative arrays, the destination must already exist as an associative array. If not, array_copy() will create it as a standard array and try to interpret the key names from the associative source as arithmetic variable names, with ugly results.
Isaac Schwabacher's solution is more robust in this regard, but it can't be tidily wrapped up in a function because its eval step evaluates an entire declare statement and bash treats those as equivalent to local when they're inside a function. This could be worked around by wedging the -g option into the evaluated declare but that might give the destination array more scope than it's supposed to have. Better, I think, to have array_copy() perform only the actual copy into an explicitly scoped destination.

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.

Answer 7

Problem is to copy array in function to be visible in parent code. This solution works for indexed arrays and if before copying are predefined as declare -A ARRAY, works also for associative arrays.
function array_copy
# $1 original array name
# $2 new array name with the same content
{
local INDEX
eval "
for INDEX in \"\${!$1[#]}\"
do
$2[\"\$INDEX\"]=\"\${$1[\$INDEX]}\"
done
"
}

Answer 8

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 9

Managed to copy an array into another.
firstArray=()
secondArray=()
firstArray+=("Element1")
firstArray+=("Element2")
secondArray+=("${firstArray[#]}")
for element in "${secondArray[#]}"; do
echo "${element}"
done

Answer 10

I've found that this works for me (mostly :)) ...
eval $(declare -p base | sed "s,base,target,")
extending the sed command to edit any switches as necessary e.g. if the new structure has to be writeable, to edit out read-only (-r).

Answer 11

I've discovered what was wrong.. My if isn't installed test is two for loops that remove excess characters from file names, and spits them out if they exist on a certain web server. What it wasn't doing was removing a trailing hyphen. So, when it tested it online for availability, they were parsed out. Because "file" exists, but "file-" doesn't.