[DFW.pm] Homework for the list, and for Oct 08 meeting

[Patrick R. Michaud]

Perl 6 solution #1, the straightforward one:

    for 1..100 {
        when $_ %% (3 & 5) { say "FizzBuzz" }
        when $_ %% 5       { say "Buzz" }
        when $_ %% 3       { say "Fizz" }
        say $_;
    }


Some explanation:

  - The Perl 6 "%%" operator returns true if the left argument is evenly 
    divisible by the right argument.

  - The "when" keyword is Perl 6's version of a "switch/case" statement; 
    when the test is true, the block following it is executed and control
    skips to the end of the enclosing block.

  - The expression "$_ %% (3 & 5)" is true if $_ is evenly divisible
    by both 3 and 5.  (The "&" operator in Perl 6 creates a junction 
    of values, the numeric bitwise-and operator in Perl 6 is "+&".)

Pm