Perl:如何将数组转换为嵌套的哈希键

问题描述：

我需要将一个简单的键列表转换成一个嵌套的散列，如下所示:

I need to convert a flat list of keys into a nested hash, as follow:

我的 $hash = {};

my $hash = {};

我的@array = qw(key1 key2 lastKey Value);

my @array = qw(key1 key2 lastKey Value);

ToNestedHash($hash, @array);

会这样做:

$hash{'key1'}{'key2'}{'lastKey'} = "Value";

$hash{'key1'}{'key2'}{'lastKey'} = "Value";

答

sub to_nested_hash {
    my $ref   = \shift;  
    my $h     = $$ref;
    my $value = pop;
    $ref      = \$$ref->{ $_ } foreach @_;
    $$ref     = $value;
    return $h;
}

说明:

将第一个值作为哈希引用
取最后一个值作为赋值
剩下的就是钥匙了.
然后创建一个对基本哈希的 SCALAR 引用.
反复:
- 取消引用指针以获取散列(第一次)或将指针自动激活为散列
- 获取密钥的哈希槽
- 并将标量引用分配给哈希槽.
- (下次将自动激活到指定的哈希值).
我们知道:
- 散列或数组的占用者只能是标量或引用.
- 引用是某种标量.(my $h = {}; my $a = [];).
- 所以，\$h->{ $key } 是对堆上标量槽的引用，可能是自动激活的.
- 嵌套散列的级别"可以自动激活为散列引用如果我们如此处理它.
- That the occupants of a hash or array can only be a scalar or reference.
- That a reference is a scalar of sorts. (my $h = {}; my $a = [];).
- So, \$h->{ $key } is a reference to a scalar slot on the heap, perhaps autovivified.
- That a "level" of a nested hash can be autovivified to a hash reference if we address it as so.
这样做可能更明确:
```
foreach my $key ( @_ ) { 
    my $lvl = $$ref = {};
    $ref    = \$lvl->{ $key };
}
```
但由于重复使用这些参考习语，我完全照原样写了那行，并在发布前对其进行了测试，没有错误.

But owing to repeated use of these reference idioms, I wrote that line totally as it was and tested it before posting, without error.

至于替代方案，以下版本更容易"(想出来)

As for alternatives, the following version is "easier" (to think up)
```
sub to_nested_hash {
    $_[0] //= {};
    my $h     = shift;
    my $value = pop;
    eval '$h'.(join '', map "->{\$_[$i]}", 0..$#_).' = $value';
    return $h;
}
```
但大约慢 6-7 倍.

But about 6-7 times slower.

Perl:如何将数组转换为嵌套的哈希键

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