raku - Perl 6's uncudled else 是语句分离的特例吗？

Question

来自 syntax doc :

A closing curly brace followed by a newline character implies a statement separator, which is why you don't need to write a semicolon after an if statement block.

if True {
    say "Hello";
}
say "world";

没关系， Why is this Perl 6 feed operator a “bogus statement”? 发生了什么| .

但是，这条规则如何适用于未拥抱的 else ?这是特例吗？

if True {
    say "Hello";
}
else {
    say "Something else";
}
say "world";

或者， with-orwith example怎么样？ :

my $s = "abc";
with   $s.index("a") { say "Found a at $_" }
orwith $s.index("b") { say "Found b at $_" }
orwith $s.index("c") { say "Found c at $_" }
else                 { say "Didn't find a, b or c" }

Answer 1

您找到的文档并不完全正确。 documentation has been updated and is now correct .现在是这样写的:

Complete statements ending in bare blocks can omit the trailing semicolon, if no additional statements on the same line follow the block's closing curly brace }.

...

For a series of blocks that are part of the same if/elsif/else (or similar) construct, the implied separator rule only applies at the end of the last block of that series.

原答案:

查看 if 的语法在 nqp 和 Rakudo ，似乎是 if/elsif/else一组块被一起解析为一个控制语句。
if 的规则在 nqp

rule statement_control:sym<if> {
    <sym>\s
    <xblock>
    [ 'elsif'\s <xblock> ]*
    [ 'else'\s <else=.pblock> ]?
}

( https://github.com/perl6/nqp/blob/master/src/NQP/Grammar.nqp#L243 ，截至 2017 年 8 月 5 日)
if 的规则位于乐堂

rule statement_control:sym<if> {
    $<sym>=[if|with]<.kok> {}
    <xblock(so ~$<sym>[0] ~~ /with/)>
    [
        [
        | 'else'\h*'if' <.typed_panic: 'X::Syntax::Malformed::Elsif'>
        | 'elif' { $/.typed_panic('X::Syntax::Malformed::Elsif', what => "elif") }
        | $<sym>='elsif' <xblock>
        | $<sym>='orwith' <xblock(1)>
        ]
    ]*
    {}
    [ 'else' <else=.pblock(so ~$<sym>[-1] ~~ /with/)> ]?
}

( https://github.com/rakudo/rakudo/blob/nom/src/Perl6/Grammar.nqp#L1450 截至 2017 年 8 月 5 日)