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
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)