perl6 – Perl 6的其他问题是否是一个特殊的声明分离案例？

从 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";

这很好,Why is this Perl 6 feed operator a “bogus statement”?发生了什么.

但是,这条规则对于一个不受约束的其他人如何运作？这是一个特例吗？

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" }

解决方法

您找到的文档并不完全正确. documentation has been updated and is now correct.它现在写着：

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.

原始答案：

查看if在nqp和Rakudo中的if语法,似乎if / elsif / else组块一起被解析为一个控制语句.

在nqp中if的规则

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)

在Rakudo的if规则

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/)> ]?
}

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

原文地址：https://www.jb51.cc/Perl/172181.html