Bash script for listing the files in a directory in a specific format witht the BASH shell

Question

I am currently working on some C++ code on a GNU/Linux system and my source-code folder is filled with .cpp files and .h files.

In general for this code, every .cpp file has a corresponding .h header file, but not necessarily vice versa. In the output below -- indicates that there is no corresponding .cpp file for the listed header file

I would like to either write a bash script to do this by say defining an extra flag in my .bashrc / .zshrc , such that the listing of the files occurs in this format. Say I have 7 files, some .cpp and some .h

$ listscript
hello1.cpp hello1.h
hello2.cpp hello2.h
   --      hello3.h 
hello4.cpp hello4.h      

Answer 1

#!/usr/bin/env bash
declare files=(*)
declare file= left= right= width=10
declare -A listed=()
for file in "${files[@]}"; do
    if [[ $file == *.h ]]; then
        continue
    elif (( ${#file} > width )); then
        width=${#file}
    fi
done
for file in "${files[@]}"; do
    if [[ ${listed[$file]} == 1 ]]; then
        continue
    elif [[ $file == *.cpp ]]; then
        left=$file right=${file%.cpp}.h
    elif [[ $file == *.h ]]; then
        left=${file%.h}.cpp right=$file
    else
        left=$file right=
    fi

    [[ $left ]]     && listed["$left"]=1
    [[ $right ]]    && listed["$right"]=1

    [[ -e $left ]]  || left='--'
    [[ -e $right ]] || right='--'

    printf "%-*s %s\n" "$width" "$left" "$right"
done