I haven't programmed in Perl in over 10 years so maybe this is something obvious to more experienced Perl programmers. I searched for an answer but didn't find anything.
我没有在Perl中编程超过10年，所以对于更有经验的Perl程序员来说这可能是显而易见的。我搜索了一个答案，但没有找到任何答案。
My question is: why are references to anonymous arrays scalar?
我的问题是：为什么引用匿名数组标量？
For example in the following code:
例如，在以下代码中：
#!/usr/bin/perl
use strict;
use feature qw(say);
my @array1 = ('one');
say 'array ref 1: ' . \@array1;
my @array2 = ('one', 'two');
say 'array ref 2: ' . \@array2;
say 'array ref 3: ' . \('one');
say 'array ref 4: ' . \('one', 'two');
exit 0;
The result is:
结果是：
array ref 1: ARRAY(0x1e1b1c0)
array ref 2: ARRAY(0x1e1b190)
array ref 3: SCALAR(0x1e1b280)
array ref 4: SCALAR(0x1e10c40)
Why are array ref 3 and array ref 4 scalar?
为什么数组引用3和数组引用4标量？
3 个解决方案
#1
14
All references are scalars. When you stringify a reference, it includes the type of what it's referencing. That means you have a reference to a scalar.
所有引用都是标量。对字符串化引用时，它包括引用的类型。这意味着你有一个标量的引用。
[ ] is the operator that constructs an array. ( ) doesn't create any arrays.
[]是构造数组的运算符。 （）不会创建任何数组。
You want
你要
say 'array ref 3: ' . ['one'];
say 'array ref 4: ' . ['one', 'two'];
Normally, parens just change precedence. If that were true here,
通常情况下，parens只会改变优先权。如果这是真的，
\('one', 'two')
would be equivalent to
相当于
('one', \'two')
That's because a comma/list operator in scalar context normally returns that to which its last expression evaluates. But \(...) is special-cased to be equivalent to the following more useful construct:
这是因为标量上下文中的逗号/列表运算符通常会返回其最后一个表达式计算的值。但\（...）是特殊的，等同于以下更有用的结构：
(\'one', \'two')
Taking a reference to an enumerated list is not the same as using square brackets--instead it's the same as creating a list of references!
引用枚举列表与使用方括号不同 - 而是与创建引用列表相同！
@list = (\$a, \@b, \%c); @list = \($a, @b, %c); # same thing!
That means that
这意味着
say 'array ref 4: ' . \('one', 'two');
is equivalent to
相当于
say 'array ref 4: ' . (\'one', \'two');
which is equivalent to
这相当于
say 'array ref 4: ' . \'two';
#2
8
From perlref:
来自perlref：
Taking a reference to an enumerated list is not the same as using square brackets--instead it's the same as creating a list of references!
引用枚举列表与使用方括号不同 - 而是与创建引用列表相同！
@list = (\$a, \@b, \%c); @list = \($a, @b, %c); # same thing!
You get a scalar reference because the item in the list that ends up with its reference passed to the LHS is a scalar.
您会得到一个标量引用，因为列表中以其引用传递给LHS的项目是标量。
#3
3
It seems as if you expect that these are equivalent:
看起来好像你期望它们是等价的：
my @array1 = ('one', 'two');
my $array2 = \('one', 'two');
But that is not the correct syntax to get an anonymous array. You need square brackets to create an anonymous array.
但这不是获取匿名数组的正确语法。您需要使用方括号来创建匿名数组。
my $array3 = ['one', 'two']; # a reference to an anonymous array
my $array4 = \['one', 'two']; # a reference to a reference to an anonymous array
#1
14
All references are scalars. When you stringify a reference, it includes the type of what it's referencing. That means you have a reference to a scalar.
所有引用都是标量。对字符串化引用时，它包括引用的类型。这意味着你有一个标量的引用。
[ ] is the operator that constructs an array. ( ) doesn't create any arrays.
[]是构造数组的运算符。 （）不会创建任何数组。
You want
你要
say 'array ref 3: ' . ['one'];
say 'array ref 4: ' . ['one', 'two'];
Normally, parens just change precedence. If that were true here,
通常情况下，parens只会改变优先权。如果这是真的，
\('one', 'two')
would be equivalent to
相当于
('one', \'two')
That's because a comma/list operator in scalar context normally returns that to which its last expression evaluates. But \(...) is special-cased to be equivalent to the following more useful construct:
这是因为标量上下文中的逗号/列表运算符通常会返回其最后一个表达式计算的值。但\（...）是特殊的，等同于以下更有用的结构：
(\'one', \'two')
Taking a reference to an enumerated list is not the same as using square brackets--instead it's the same as creating a list of references!
引用枚举列表与使用方括号不同 - 而是与创建引用列表相同！
@list = (\$a, \@b, \%c); @list = \($a, @b, %c); # same thing!
That means that
这意味着
say 'array ref 4: ' . \('one', 'two');
is equivalent to
相当于
say 'array ref 4: ' . (\'one', \'two');
which is equivalent to
这相当于
say 'array ref 4: ' . \'two';
#2
8
From perlref:
来自perlref：
Taking a reference to an enumerated list is not the same as using square brackets--instead it's the same as creating a list of references!
引用枚举列表与使用方括号不同 - 而是与创建引用列表相同！
@list = (\$a, \@b, \%c); @list = \($a, @b, %c); # same thing!
You get a scalar reference because the item in the list that ends up with its reference passed to the LHS is a scalar.
您会得到一个标量引用，因为列表中以其引用传递给LHS的项目是标量。
#3
3
It seems as if you expect that these are equivalent:
看起来好像你期望它们是等价的：
my @array1 = ('one', 'two');
my $array2 = \('one', 'two');
But that is not the correct syntax to get an anonymous array. You need square brackets to create an anonymous array.
但这不是获取匿名数组的正确语法。您需要使用方括号来创建匿名数组。
my $array3 = ['one', 'two']; # a reference to an anonymous array
my $array4 = \['one', 'two']; # a reference to a reference to an anonymous array