perl6 – Perl 6的其他问题是否是一个特殊的声明分离案例?

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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nqpRakudo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)

原文链接:https://www.f2er.com/Perl/172181.html

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