分配给列表容器混淆

p6s*_*eve 11 raku

我可能患有大脑衰退,但根据有关项目和列表分配的文档(https://docs.raku.org/language/variables#Item_and_list_assignment),

分配给列表容器(列表上下文)总是触发列表分配。

然而,这似乎与我从代码中得到的内容相冲突(这里在 raku repl 中复制)。

> my %syns-by-name = %(Bq => ["Bq", "becquerel", "becquerels"], C => ["C", "coulomb", "coulombs"],)
    {Bq => [Bq becquerel becquerels], C => [C coulomb coulombs]}
> my Str @res = %syns-by-name{'Bq'};
    Type check failed in assignment to @res; expected Str but got Array (["Bq", "becquerel", ...) in block <unit> at <unknown file> line 1
> my Str @res = [|%syns-by-name{'Bq'}];
    [Bq becquerel becquerels]
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这是一个错误,还是我误解了意图是什么......?

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rai*_*iph 7

TL;DR如果非本地项通过赋值可变,则它是Scalar. 列表分配不会展平Scalar项目。[1]

一个暗示

考虑这个代码:

my %map = :a;
%map<a> = 42;
say %map<a>; # 42
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该作业有效,因为:

say %map<a>.VAR.WHAT; # (Scalar)
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“列表分配”

考虑这个代码:

my $scalar = 1,2;       # Useless use of constant integer 2
say $scalar;            # 1

my @list1 = 1,2;        # "list assignment", so RHS is iterated
say @list1;             # [1 2]
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所以这是项目和列表分配之间的一个区别。

my @list3 = [1,2];      # Again, RHS is iterated
say @list3;             # [1 2]

my @list2 = (1,2);      # Again, RHS is iterated
say @list2;             # [1 2]

my @list4 = 1,(2,3);    # Again, RHS is iterated. Second element stays as a `List`.
say @list4;             # [1 (2 3)]

my @list5 = 1,[2,3];    # Again, RHS is iterated. Second element stays as an `Array`.
say @list5;             # [1 [2 3]]
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如果 RHS 上只列出了一项,并且它不是Scalar,则列表分配会将其展平。但在所有其他情况下,列表分配不会使项目变平。

my @list6 = $[1,2];     # RHS is a `Scalar`. So it doesn't get flattened.
say @list6;             # [[1 2]]
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我好糊涂!

Q打高尔夫球的情况:

my Str @res = %( :a[42,99] )<a>;
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这会产生同样的错误。

因为:

say .VAR.WHAT given :a[42,99]<a>;       # (Array)
say .VAR.WHAT given (% = :a[42,99])<a>; # (Scalar)
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脚注

[1]当猜测产生惊喜,而您将其转化为学习时,您就意识到并理想化了对ERN的投资。