bash - Add an element before each item on array

Question

I have an array containing strings to exclude with grep from the output of another program. I need to add an -e before each element. For instance:

exclude=("$0" /usr/sbin/crond)
needs-restarting | grep -Fwiv "${exclude[@]}"

Now I know in this case I could prepend --regexp= (or just -e) to each element like so: exclude=( "${exclude[@]/#/--regexp=}" )

But in the general case, how would I go about it? I came up with this but maybe there's a simpler way.

i=0
for elem in "${exclude[@]}"; do
  exclude[i]='-e'
  exclude[i+1]="$elem"
  ((i+=2))
done
declare -p exclude

Answer 1

In bash 4.4+, you could do:

readarray -td '' array < <(
  ((${#array[@]})) && printf -- '-e\0%s\0' "${array[@]}"
)

Here using \0 as the delimiter as bash variables can't contain NUL bytes anyway. If you know the array is not going to be empty, you can skip the ((${#array[@]})) &&.

Example:

• before:

  bash-5.0$ array=($'a\nb' '' 'c d' e)
  bash-5.0$ typeset -p array
  declare -a array=([0]=$'a\nb' [1]="" [2]="c d" [3]="e")
  

• after:

  bash-5.0$ typeset -p array
  declare -a array=([0]="-e" [1]=$'a\nb' [2]="-e" [3]="" [4]="-e" [5]="c d" [6]="-e" [7]="e")
  

In zsh, you could use its array zipping operator:

opt=-e
(($#array == 0)) || array=("${(@)opt:^^array}")

Or this convoluted one:

set -o extendedglob # for (#m)
array=("${(Q@)"${(@z)array//(#m)*/-e ${(qq)MATCH}}"}")

Where we replace each element with -e <the-element-quoted> (with the qq flag), and then use z to parse that quoting back into a list of elements (where -e and <the-element-quoted> are then separated out), and Q to remove the quotes (and @ within quotes used to preserve empty elements if any).

Answer 2

You can simplify your proposed loop a little:

exclude_args=()
for elem in "${exclude[@]}"; do
  exclude_args+=('-e' "$elem")
done
exclude=("${exclude_args[@]}")   # Optional, if you want to replace the original array's contents