in reply to printing 20 characters in a line.

my $nl = 1;
for my $i (0..$#array) {
   if (!$nl && $i%20==0) {
      print("\n$array[$i]");
      $nl = 1;
   } else {
      print(" $array[$i]");
      $nl = 0;
   }
}
print("\n") if !$nl;

[download]

Replies are listed 'Best First'.
Re^2: printing 20 characters in a line.
by heidi (Sexton) on Nov 14, 2008 at 05:11 UTC
    I got it... how is this one??? 
    #!/usr/bin/perl
my $str="30 30 36 33 36 36 36 33 36 34 26 24 21 17 17 23 27 27 31 31 3
+4 33 33 33 36 36 37 38 38 38 38 38 35 40 34 34 34 31 37 36 37 37 40 4
+0 49 40 40 40 40 40 40 40 40 33 30 30 30 30 30 42 45 45 45 45 45 49 4
+2 42";
@s=split(/ /,$str);
for($i=1;$i<=scalar(@s);$i++) {
print "$s[$i] ";
if($i%20==0) {
print "\n";
}
}
print "\n";

    [download]
    simple uh.. y didnt i think abt it when i tried first :)
[reply]
[d/l]
      • You skip the first element.
      • You print an extra element (undef) at the end. use warnings; would have caught that.
      • The use of scalar there is unnecessary.
      • You really should use use strict;.
      • for (my $i=0; $i<@s; $i++)
        is a rather complex way of saying
        for my $i (0..$#s)

      I had a bug in mine too (leading space) 

      #!/usr/bin/perl
use strict;
use warnings;
my $str = "30 30 36 33 36 36 36 33 36 34"
        ." 26 24 21 17 17 23 27 27 31 31"
        ." 34 33 33 33 36 36 37 38 38 38"
        ." 38 38 35 40 34 34 34 31 37 36"
        ." 37 37 40 40 49 40 40 40 40 40"
        ." 40 40 40 33 30 30 30 30 30 42"
        ." 45 45 45 45 45 49 42 42";
my @str = split(/ /, $str);
my $nl = 1;
for my $i (0..$#str) {
   if    ($i == 0     ) { $nl = 0;              }
   elsif ($i % 20 == 0) { $nl = 1; print("\n"); }
   else                 { $nl = 0; print(" ");  }
   print($str[$i]);
}
print("\n") if !$nl;

      [download]

      Tested.

[reply]
[d/l]
[select]

In Section Seekers of Perl Wisdom