Bash - draw a vertical line behind lines with variable length

You could print the lines without alignment and format the output with column -t and a dummy delimiter character:

#!/bin/bash

while read -r line; do
  if [ -z "$line" ]; then
    echo
    continue
  fi
  printf '%[email protected]| %%%s\n' "$line" "$((++n))"
done < file | column -e -s'@' -t | sed 's/ |/|/'

Here, I added a @ as dummy character before the | indicating the end of the column. The sed command at the end is used to remove one additional space character before the |. Option -e is needed to keep empty lines in the output.

Output:

c4-1 d e c            | %1
c d e c               | %2
e-2 f g2              | %3
e4 f g2               | %4
g8-4\( a-5 g f\) e4 c | %5
g'8\( a g f\) e4 c    | %6
c-1 r c2              | %7
c4 r c2               | %8

Using awk + GNU wc assuming all characters in the input are single-width:

$ awk -v f="$(wc -L < ip.txt)" '{printf "%-*s | %%%s\n", f, $0, NR}' ip.txt
c4-1 d e c            | %1
c d e c               | %2
e-2 f g2              | %3
e4 f g2               | %4
g8-4\( a-5 g f\) e4 c | %5
g'8\( a g f\) e4 c    | %6
c-1 r c2              | %7
c4 r c2               | %8

Plain bash: works with bash version >= 4.0

#!/bin/bash
mapfile -t lines < file
max=0
for line in "${lines[@]}"; do
    max=$(( ${#line} > max ? ${#line} : max ))
done
for i in "${!lines[@]}"; do
    printf "%-*s | %%%d\n" $max "${lines[i]}" $((i+1))
done

For older bash versions, replace mapfile with a while-read loop: this works with version 3.2

#!/bin/bash
lines=()
max=0
while IFS= read -r line || [[ -n "line" ]]; do
    lines+=("$line")
    max=$(( ${#line} > max ? ${#line} : max ))
done < file
for i in "${!lines[@]}"; do
    printf "%-*s | %%%d\n" $max "${lines[i]}" $((i+1))
done