Yes you are right.




When the shell script starts, both arg1 and arg2 have the same value.

So I use the counter with value 0, to determine this case, and advance the second for-loop to the next argument, and from there, I check if arg2=arg1.


If the second loop ends, the first loop progresses to the next argument.

The second loop restarts again with the first argument, and proceeds until it meets the first loop, where arg2=arg1 and counter -eq 0.

arg2 then proceeds to the next argument, until it ends, if no duplicate arguments found.

Then the first loop progresses to the next argument. etc.

The code that works, is the first I posted, and is this:

Code:

#!/bin/sh

for arg1 in "$@"
do
    counter=0


   for arg2 in "$@"
   do

     if [ "$arg2" = "$arg1" ] && [ $counter -eq 0 ]
     then
       counter=$((counter+1))

       continue 
    fi


   if [ "$arg2" = "$arg1" ]
   then
    
     echo
     echo "Error: Two or more arguments are the same."
     echo
     echo "Exiting..."
     echo

     exit 1
    fi

  done

done

exit 0