为什么这个 Perl 6 feed 操作符是一个“虚假声明"? [英] Why is this Perl 6 feed operator a "bogus statement"?
问题描述
我从 第 10 天 - Feed 操作员中获取了这个示例Perl 6 2010 Advent Calendar 中的 .uc 略有更改，.ucfirst 不再存在:
我的@rakudo-people = ;@乐堂人==>grep {/at/} ==>地图 { .uc } ==>我的@who-it's-at;说~@who-it's-at; 我用一些额外的空格来写它:
我的@rakudo-people = ;@乐堂人==>grep {/at/}==>地图 { .uc } ==>我的@who-it's-at;说~@who-it's-at; 现在是虚假陈述":
<块引用>===SORRY!=== 编译时出错...虚假陈述------>==>grep {/at/}⏏期待以下任何一项:后缀字首语句结束学期 仅此示例就没有问题.当前文档中的一些示例可以表现出相同的行为.
如果我在违规行的末尾添加 unspace，它会起作用再次:
我的@rakudo-people = ;@乐堂人==>grep {/at/} \==>地图 { .uc } ==>我的@who-it's-at;说~@who-it's-at; 奇怪的是，该行末尾的注释不起作用.我原以为它会吃掉有问题的空白.
feed 运营商 说:
<块引用>对于采用单个参数或第一个参数是块的例程/方法，通常需要使用括号进行调用
那行得通:
我的@rakudo-people = ;@乐堂人==>grep( {/at/} )==>地图 { .uc } ==>我的@who-it's-at;说~@who-it's-at; 但是为什么在第一种形式中这不是问题?空白在这里做什么?经常需要"包括哪些情况?
来自关于分隔语句的 Perl 6 文档:
<块引用>右花括号后跟换行符表示语句分隔符
换句话说，只要块的右大括号是一行中的最后一个东西(不包括空格和注释)，解析器就会在它后面隐式插入一个分号.
这允许我们在 } 后不加分号的情况下编写如下内容:
my @foo = @bar.map: {...}分号在那里会很丑，如果你先把循环写成 for @bar { ... } 然后决定把它变成一个 map像这样的分配，添加尾随分号会很烦人，很容易忘记.所以在大多数情况下，这种在行尾块之后自动终止语句是有帮助的.
但是，喂食块的饲料运营商是一种(可能是唯一一种)让人们绊倒的情况.
为了防止它发生，插入 \(又名 unspace) 在块之后，正如您已经注意到的.unspace 使包含换行符的空白对解析器不可见，因此不会应用上述基于换行符的解析器规则.
I took this example from Day 10 – Feed operators of the Perl 6 2010 Advent Calendar with the slight change of .uc for the .ucfirst that's no longer there:
my @rakudo-people = <scott patrick carl moritz jonathan jerry stephen>;
@rakudo-people
==> grep { /at/ } ==> map { .uc } ==> my @who-it's-at;
say ~@who-it's-at;
I write it slightly differently with some additional whitespace:
my @rakudo-people = <scott patrick carl moritz jonathan jerry stephen>;
@rakudo-people
==> grep { /at/ }
==> map { .uc } ==> my @who-it's-at;
say ~@who-it's-at;
Now it's a "bogus statement":
===SORRY!=== Error while compiling ... Bogus statement ------> ==> grep { /at/ }⏏<EOL> expecting any of: postfix prefix statement end term
This isn't a problem with just this example. Some examples in the current docs can exhibit the same behavior.
If I add an unspace to the end of the offending line, it works again:
my @rakudo-people = <scott patrick carl moritz jonathan jerry stephen>;
@rakudo-people
==> grep { /at/ } \
==> map { .uc } ==> my @who-it's-at;
say ~@who-it's-at;
Curiously, a comment at the end of that line does not work. I would have thought it would have eaten the offending whitespace.
The feed operator says:
In the case of routines/methods that take a single argument or where the first argument is a block, it's often required that you call with parentheses
That works:
my @rakudo-people = <scott patrick carl moritz jonathan jerry stephen>;
@rakudo-people
==> grep( { /at/ } )
==> map { .uc } ==> my @who-it's-at;
say ~@who-it's-at;
But why wasn't that a problem in the first form? What is the whitespace doing here? And what situations are included in "often required"?
From the Perl 6 documentation on Separating Statements:
A closing curly brace followed by a newline character implies a statement separator
In other words, whenever the closing brace of a block is the last thing in a line (not counting whitespace and comments), then the parser implicitly inserts a semicolon after it.
This allows us to write things like the following without a semicolon after the }:
my @foo = @bar.map: {
...
}
The semicolon would be ugly there, and if you had first written the loop as for @bar { ... } and then decided to turn it into a map with assignment like this, adding the trailing semicolon would be annoying and easy to forget. So for the most part, this automatic statement termination after end-of-line blocks is helpful.
However, feed operators feeding into blocks are one case (possibly the only one) where it trips people up.
To prevent it from happening, insert \ (a.k.a. unspace) after the block, as you've already noted. The unspace makes the whitespace which includes the newline invisible to the parser, and thus the aforementioned newline-based parser rule won't be applied.
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