How to add a space after special characters in bash script?

Question

I have a text file with something like,

!aa
@bb
#cc
$dd
%ee

expected output is,

! aa
@ bb
# cc
$ dd
% ee

What I have tried, echo "${foo//@/@ }".

This does work fine with one string but it does not work for all the lines in the file. I have tried with this while loop to read all the lines of the file and do the same using echo but it does not work.

while IFS= read -r line; do
    foo=$line
    sep="!@#$%"
    echo "${foo//$sep/$sep }"
done < $1

I have tried with awk split but it does not give the expected output. Is there any workaround for this? by using awk or sed.

Answer 1

The following assumes you want to add a space after every character in the !@#$% set (even if it is the last character in a line). Test file:

$ cat file.txt
a!a
@bb
c#c
$dd
ee%
foo
%b%r
$ sep='!@#$%'

With sed:

$ sed 's/['"$sep"']/& /g' file.txt
a! a
@ bb
c# c
$ dd
ee% 
foo
% b% r

With awk:

$ awk '{gsub(/['"$sep"']/,"& "); print}' file.txt
a! a
@ bb
c# c
$ dd
ee% 
foo
% b% r

With plain bash (not recommended, it is too slow):

$ while IFS= read -r line; do
    str=""
    for (( i=0; i<${#line}; i++ )); do
      char="${line:i:1}"      
      str="$str$char"
      [[ "$char" =~ [$sep] ]] && str="$str "
    done
    printf '%s\n' "$str"
  done < file.txt
a! a
@ bb
c# c
$ dd
ee% 
foo
% b% r

Or (not sure which is the worst):

$ while IFS= read -r line; do
    for (( i=0; i<${#sep}; i++ )); do
      char="${sep:i:1}"
      line="${line//$char/$char }"
    done
    printf '%s\n' "$line"
  done < file.txt
a! a
@ bb
c# c
$ dd
ee% 
foo
% b% r