sub compare {
  $a cmp $b
}


sub binsearch {
   # Returns the index of the position before
   # the one where $s would be found, when $s
   # is not found.

   my ($f, $s, $list) = @_;

   my $i = 0;
   my $j = $#$list;
   my $k;
   my $c;

   $a = $s;

   for (;;) {
      $k = int(($j-$i)/2) + $i;

      $b = $list->[$k];
      $c = &$f();

      return $k if ($c == 0);

      if ($c < 0) {
         $j = $k-1;
         return $j if ($i > $j);
      } else {
         $i = $k+1;
         return $k if ($i > $j);
      }      
   }
}


@a = qw( z b x c y a );
@b = qw( o m n );

@a = sort compare @a;
@b = sort compare @b;

splice(@a, binsearch(\&compare, $b[0], \@a)+1, 0, @b);

print(@a);  # abcmnoxyz
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Of course, in your case, it will be
splice(@known, binsearch(\&byIndex, $newValue, \@known)+1, 0, $newValue);

Update: Passing array to binsearch using a reference now. oops!

Passing @a not \@a means that perl copies the entire array to the function for no reason, and thus loses a lot of the potential efficiency.

This is the fastest, especially on a 10,000 record file.
Oh I wanted to give thanks to all of you who helped me earlier on Figuring out how to embed c into PERL. Still not perfect at it, but i got it to work for what I need. Thanks Again to all you Wise monks. Ketema

Yes you can do a binary search. Yet is will find the correct placement faster for large lists. In fact I have a module on cpan called File::SortedSeek that implements binary searches in large files so I appreciate the algorithm :-) However in this scenario having found the index you then have a practical problem. It is not possible to insert a value into the middle of an array, which is really just a contiguous sequence of SV* What you have to do is make space for it. Splice does that by moving a large chunk (or the entire array). The bubble algorithm also does it and is (in past testing) about twice as fast as just calling sort. Here is a binary search implementation. Benchmarking the options is left to you.

my @ary = (1..4, 6..10);

for my $val( 0, 5, 11, 0, 5, 11 ) {
    binary( \@ary, $val );
    print "@ary\n";
}

sub binary {
    my ( $ary, $val ) = @_;
    my ( $min, $max, $last, $i ) = ( 0, scalar @{$ary}, 0, 0 );
    while ( 1 ) {
        $i = int( ($min+$max)/2 );
        # print "i=$i\n";  
      last if $last == $i;
        $last = $i;
        if ( $ary->[$i] < $val ) {
            $min = $i;
        }
        elsif ( $ary->[$i] > $val ) {
            $max = $i;
        }
        else {
            # values are equal so we have a valid index
            last;
        }
    } 
    $i++ if $i;   # for index 0 we want that, otherwise we want next p
+osition   
    splice @$ary, $i, 0, $val;
}

__DATA__
0 1 2 3 4 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9 10 11
0 0 1 2 3 4 5 6 7 8 9 10 11
0 0 1 2 3 4 5 5 6 7 8 9 10 11
0 0 1 2 3 4 5 5 6 7 8 9 10 11 11
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