CODEWITHSUNDEEP
perl
hanktank45
hanktank45

Reputation: 19

Perl counter print list with commas except last item

I am encountering an issue while trying to execute a perl script that takes 2 number arguments, well say $ARGV[0] is 2, and $ARGV[1] is 4. I need to print a list that shows 2,3,4 with no comma after the last item. Below is my script as it is now:

unless ((@ARGV)==2){
        print "error: incorrect number of arguments",
        "\n",
        "usage: inlist.pl a b (where a < b)",
        "\n";
                exit VALUE;
}

if ($ARGV[0] > $ARGV[1]){
        print "error: first argument must be less than second argument",
        "\n",
        "usage: intlist.pl a b (where a < b)",
        "\n";
                exit VALUE;
}
else {

        $COUNTER=$ARGV[0];

        while($COUNTER <= $ARGV[1]){
 print $COUNTER;
                $COUNTER += 1;
        if ($COUNTERELATIONAL < $ARGV[1]){
                print ", ";
        }
        else {
                print "\n";
        }
$COUNTERSYNTAX
        }
}
exit VALUE;

I tried using join but to no avail I keep getting as return of 2,3,4,

I feel like i must be missing something simple

Upvotes: 0

Views: 166

Answers (2)

Dave Cross
Dave Cross

Reputation: 69314

Rewriting your code to simplify it:

# Prefer 'if' over 'unless' in most circumstances.
if (@ARGV != 2) {
    # Consider using 'die' instead of 'print' and 'exit'.
    print "error: incorrect number of arguments\n",
          "usage: inlist.pl a b (where a < b)\n";

    # Not sure what VALUE is, but I assume you've
    # defined it somewhere.
    exit VALUE;
}

if ($ARGV[0] > $ARGV[1]) {
    # Consider using 'die' instead of 'print' and 'exit'.
    print "error: first argument must be less than second argument\n",
          "usage: intlist.pl a b (where a < b)\n";

    exit VALUE;
}

# Removed 'else' branch as it's unnecessary.
# Use 'join' instead of a complicated loop.
print join(',', $ARGV[0] .. $ARGV[1]), "\n";

# This looks like a successful execution to me, so
# that should probably be 'exit 0'.
exit VALUE;

If I was writing it for myself, I'd make it a little shorter:

my %errors = (
  NUMBER => 'incorrect number of arguments',
  RANGE  => 'first argument must be less than second argument',
);

my $usage = 'usage: inlist.pl a b (where a < b)';

die "$errors{NUMBER}\n$usage\n" if @ARGV != 2;
die "$errors{RANGE}\n$usage\n"  if $ARGV[0] > $ARGV[1];

print join(',', $ARGV[0] .. $ARGV[1]), "\n";

exit 0;

Upvotes: 2

hanktank45
hanktank45

Reputation: 19

I figured it out:

while($COUNTER <= $ARGV[1]){
        print $COUNTER;
        $COUNTER += 1;
if ($COUNTER <= $ARGV[1]){
        print ", ";
}
else {
        print "\n";
}

I needed to change the if to $COUNTER and <= and it printed correctly. Thank you for the join suggestion, that would have worked if I had designed the script more efficiently

Upvotes: 0

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