Question

我正在尝试使用Enumeration role in Perl 6的示例（作为解决文档问题Enumeration role is not documented的一部分）。我想出了一个简单的例子：

text = "I am happy with <body> <h1>This is a Heading</h1>  <p>This is a paragraph.</p> </body> 3333  <body> <h1>This is a Heading</h1>  <p>This is a paragraph.</p> </body> your code"
print(remove_tags(text))
>
I am happy with  3333   your code

text = "x<=1 <div> cookies </div>"
print(remove_tags(text))
>
x<=1 

text = "I am <a happy with <body> </body> lal"
print(remove_tags(text))
>
I am <a happy with  lal

理论上，class DNA does Enumeration { my $DNAindex = 0; my %pairings = %( A => "T", T => "A", C => "G", G => "T" ); method new( $base-pair where "A" | "C" | "G" | "T" ) { self.bless( key => $base-pair, value => %pairings{$base-pair}, index => 33); } multi method gist(::?CLASS:D:) { return "$!key -> $!value with $!index"; } } for <A C G T>.roll( 16 ) -> $letter { my DNA $base = DNA.new( $letter ); say "Pairs ", $base.pair, " with ", $base.gist; } Enumeration has，并且我尝试使用$!index为其分配一个值；但是，它返回的只是类似

index => 33

任何其他直接为$！index赋值的方法，我得到the "cannot assign to an immutable value，我遇到了另一个问题。根据{{3}}，这可能是一个错误；在这种情况下，我想知道是否有解决方法。

Answer 1

这是一个错误Cannot change native role attribute from consuming class（与您链接的答案中提到的错误无关）。

我不知道解决方法。

为混合角色的属性分配值

1 个答案: