Linux Cut Command – Does Cut Return Any Fields If Separator Does Not Exist

As others have pointed out, you should not parse the output of ls. Assuming you are using ls only as an example and will be parsing something else, there are a few ways of doing what you want:

1. cut with -d and -f

   cut -d ' ' -f 1,2,3,4,9
   

   from man cut:

   -d, --delimiter=DELIM
         use DELIM instead of TAB for field delimiter
   
   -f, --fields=LIST
         select only these fields;  also print any line
         that contains no delimiter  character,  unless
         the -s option is specified
   

   Specifically for ls this is likely to fail since ls will change the amount of whitespace between consecutive fields to make them align better. cut treats foo<space>bar and foo<space><space>bar differently.

2. awk and its variants split each input line into fields on white space so you can tell it to print only the fields you want:

   awk '{print $1,$2,$3,$4,$9}'
   

3. Perl

   perl -lane 'print "@F[0 .. 3,8]"'
   

You could use var=$(echo "${versionfull}.0" | cut -d'.' -f3).

In the first case, versionfull will contain 6.40.4.0, ignoring the padding and returning 4 as needed. In the second case, the .0 will be padded and returned.