How can I take a sub-array in bash of the first N elements of a string array with elements containing spaces?

Question

This question is similar to this one but it differs from it:

Consider this array with string elements which may contain spaces:

a@W:$ arr=("eins" "zwei" "eins plus zwei" "vier" "fünf")
a@W:$ echo ${#arr[@]}   # ok, 5 elements, as expected
5
a@W:$ echo ${arr[3]}    # the fourth one, indexing starts at 0
vier
a@W:$ echo ${arr[2]}    # the third one, which contains two blancs
eins plus zwei
a@W:$ ar2=${arr[@]:0:3} # I only want the firs three of them
a@W:$ echo ${ar2[@]}
eins zwei eins plus zwei
a@W:$ echo ${#ar2[@]}   # but they are all glued together into one element
1
a@W:$ 

What mus I do to prevent this gluing them all together? The string containing spaces, "eins plus zwei" shall stay the third element.

Answer 1

You are assigning the array slice to a non-array variable, so you are getting a string. If you instead assign to an array, and quote it to protect from split+glob, it will work as you expect:

$ arr=("eins" "zwei" "eins plus zwei" "vier" "fünf")
$ ar2=( "${arr[@]:0:3}" )
$ echo "${ar2[@]}"
eins zwei eins plus zwei
$ echo ${#ar2[@]}
3

You can even watch this happen in real time with set -x:

$ set -x
$ ar2=${arr[@]:0:3}
+ ar2='eins zwei eins plus zwei'

As you can see, the expansion of ${arr[@]:0:3} is simply a string and it is being assigned to the variable passed as a single quoted string. If you want this to be split into an array, you need to process it accordingly:

$ ar2=( "${arr[@]:0:3}" )
+ ar2=(${arr[@]:0:3})