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• C3 method resolution order < 2005-11-25 17:36:06 >
  一种方法调用顺序的解决方案
  
• m:nth(19)/Perl6Examples/ < 2005-06-06 22:02:01 >
  规则修饰符/ Modifiers
  
• my $queen; rule me {Perl6|Examples18} < 2005-06-06 12:26:32 >
  Perl6 新的语法规则
  
• my enum Example17 <<:Perl(6) Today>> < 2005-06-05 12:14:08 >
  Today is 7. :)
  
• my subtype Examples16 of Perl6 where /^kissu$/ < 2005-06-04 23:59:29 >
  Perl6 中的子类型/Subtype
  
• has $.Examples14 handles 'Perl6'; < 2005-06-02 01:38:59 >
  Perl6 OO 中的 Delegation
  
• role Example13 { # Perl6 } < 2005-05-31 18:22:03 >
  Perl6's role: OO 代码复用
  
• $Perl6 = $class.bless(:Example<12>); < 2005-05-31 11:38:46 >
  Class 的构建和初始化
  
• class Examples11 is Perl6 { ... } < 2005-05-30 13:31:31 >
  Perl 6 的 OO 方式
  
• %Examples{'IO'} = open('Perl6'); < 2005-05-30 00:24:01 >
  the IO part of Perl6
  
• Perl6 | Examples & 9 ^ junction < 2005-05-26 12:35:29 >
  fantastic junction
  
• let @Examples[8] = 'Perl6'; < 2005-05-24 13:05:58 >
  let state temp
  
• sub infix:<(7)> ($Perl6, @Examples) < 2005-05-24 02:16:42 >
  Something about Operator overloading
  
• multi sub Examples (6, Perl6) {...} < 2005-05-23 22:43:33 >
  the "multi" sub
  
• sub Perl6 (*@Examples[5] is copy) < 2005-05-22 15:37:26 >
  Notes on sub of Perl6
  
• say q:2 '@*Examples.[4] &Perl6()'; < 2005-05-22 12:24:12 >
  Notes about Variable
  
• my Perl6 @Examples[3] < 2005-05-21 13:53:16 >
  Note about Perl 6 Type
  
• @Perl6 <== @Examples xx 2 < 2005-05-20 18:53:54 >
  Perl notes about Variables && Operators
  
• given @Examples[1] when Perl6 < 2005-05-19 22:05:56 >
  Perl6 "given"
  
• @Examples[0] is Perl6 < 2005-05-19 20:26:08 >
  Examples with Perl6
  
• Perl 6 Apocalypse < 2005-02-28 12:30:10 >
  阅读Larry的启示录