How to avoid importing default symbols when using a module?

Question

I am going to use Hash::Merge as an example. Consider:

use v6;
use Hash::Merge; # <-- imports all symbols marked with "is export" from Hash::Merge
my %hash1 = a1 => [1, 2, 3], b => "xxx", c => { ca => 1 }, e => 5;
my %hash2 = a1 => [1, 5, 3], b => "yyyy", c => { ca => 5, f => "a" }, d => 4;
my %res = merge-hash(%hash1, %hash2, :no-append-array);

Suppose I do not want to pollute my name space when using a module (here Hash::Merge is used as an example). I could achive this in Perl 5 by specifying an empty argument list to use:

use Hash::Merge ();   # <-- No symbols will be imported into the current namespace

Then I would call the sub routine merge-hash using its fully qualified name: Hash::Merge::merge-hash.

According to this bug report it seems like this is not possible in Perl 6. Is this correct?

Answer 1

要加载模块而不导入，请need改用：

need Hash::Merge;

对于所涉及的模块，它不会声明使用其导出的内容our，不幸的是，这意味着将其调用为：

Hash::Merge::merge-hash(...)

由于未安装在软件包中，因此无法使用。但是，仍然可以手动从导出中挖掘符号：

need Hash::Merge;
say Hash::Merge::EXPORT::DEFAULT::merge-hash({ a => 1 }, { b => 2 })

并且，为方便起见，可以使用别名：

need Hash::Merge;
my constant &merge-hash = &Hash::Merge::EXPORT::DEFAULT::merge-hash;
say merge-hash({ a => 1 }, { b => 2 });

在的基础上有一种推测的语法use Hash::Merge :MY<&merge-hash>，在当前的Perl 6版本中未实现，但是可能具有与constant此处显示的技巧相同的语义。

Answer 2

一种简单的解决方法是将模块的使用放在一个块中。

{ use Hash::Merge }

由于{}定义了作用域，因此无法逃脱。

您可以将其放置在一个do块中，以使某些东西可以逃脱。

do { use Hash::Merge }

然后，您可以做的就是将其保存下来，以便将您关心的值存储在正确的位置。

my &merge-hash = do { use Hash::Merge; &merge-hash }

my (&merge-hash,&merge-hashes) = do { use Hash::Merge; (&merge-hash, &merge-hashes) }

另一种选择是将其放置在尽可能小的范围内。

my %a = a => 1;
my %b = b => 2;

my %c;
{
  use Hash::Merge;
  %c := merge-hash %a, %b
}

要么

my %c := do {
  use Hash::Merge;
  merge-hash %a, %b
}

（:=仅使用了绑定运算符，因为的结果merge-hash已经是哈希值。）