Hi All,
I would like to do a loop on all the files with extension .out in my directory this files should be sorted alphabetically.
Then I need to assign to each of them a progressive ID of three digits (maybe in HEX format?).

First I tried to list manually the files as

ARRAY=( 
A-001.out  B-006.out ...
)
LIST=(
01 02 ...)

ELEMENTS=${#ARRAY[@]}  
echo $ELEMENTS         
# echo each element in array 
# for loop             
for (( i=0;i<$ELEMENTS;i++)); do
    echo ${ARRAY[${i}]} ${LIST[${i}]}
done

Something like this?

ls | awk '{printf("%s_%03s\n",$0, ++i)}'

This something I wrote that does the job, but there may be a smarter way

for I in `ls ${PATH}/*.out | cut -c${LENGTH}-`;
do
 ARRAY[$COUNTER]=$I
 if [ $COUNTER -lt 10 ]; then
     LIST[$COUNTER]=000$COUNTER
 else
     if [ $COUNTER -lt 100 ]; then
         LIST[$COUNTER]=00$COUNTER
     else
         if [ $COUNTER -lt 1000 ]; then
              LIST[$COUNTER]=0$COUNTER
         else
             if [ $COUNTER -lt 10000 ]; then
                  LIST[$COUNTER]=$COUNTER
             fi
         fi
     fi
 fi

fraklins codes look, easier and smart,
posting the expected output along with question is always, better than
explaining the problems in words

for f in *.out; do
  printf "%03d %s\n" $((++i)) "$f"
done

for f in *.out; do
  printf "%03X %s\n" $((++i)) "$f"
done

for f in *.out; do
  printf "%s %03X\n" "$f" $((++i))
done

cat -n <(ls -1 *.out)

thank you all