为什么@array ~~ LIST返回false,即使@array包含与LIST相同的元素？

Question

我有

@a = (1,2,3); print (@a ~~ (1,2,3))

和

@a = (1,2,3); print (@a == (1,2,3))

第一个是我期望的工作,但它不会打印任何东西.第二个打印1.

为什么？是不是智能匹配运营商~~应该匹配的情况@a ~~ (1,2,3)？

Answer 1

一秒钟,让我们考虑一下略有不同

\@a ~~ (1,2,3)

~~ 在标量上下文中计算其参数,因此上面的内容与

scalar(\@a) ~~ scalar(1,2,3)

• \@a(在任何上下文中)返回对引用的引用@a.
• 1, 2, 3在标量上下文中类似于do { 1; 2; 3 },返回3.

所以减去几个警告*,上面相当于

\@a ~~ 3

你真正想要的是什么

\@a ~~ do { my @temp = (1,2,3); \@temp }

这可以缩短为

\@a ~~ [ 1,2,3 ]

最后,~~允许\@a被写成的魔力@a,这样可以进一步缩短

@a ~~ [ 1,2,3 ]

* - 始终使用use strict; use warnings;!

Answer 2

如果您在右侧使用数组或数组引用 - 但不是列表,智能匹配会尝试执行我认为您期望的操作.

$ perl -E '@a = (1, 2, 3); say (@a ~~ (1, 2, 3))'

$ perl -E '@a = (1, 2, 3); say ((1, 2, 3) ~~ @a)' # also misguided, but different
1
$ perl -E '@a = (1, 2, 3); say (@a ~~ [1, 2, 3])'
1

Answer 3

从列表和数组有什么区别？在perlfaq.它专门向您展示您的价值选择是如何错误的.

您也可以首先写出您希望每个人工作或不工作的原因,以便我们可以纠正您的期望.为什么你认为你会得到你期望的结果？

至于智能匹配位,没有规则ARRAY ~~ LIST.智能匹配仅适用于perlsyn中其表中枚举的对.它将迫使它成为其中一对.

当您遇到这些问题时,请尝试更多案例:

#!perl
use v5.10.1;
use strict;
use warnings;

my @a = (1,2,3);
say "\@a is @a";
say "\@a ~~ (1,2,3) is ", try( @a ~~ (1,2,3) );
say "\@a ~~ [1,2,3] is ", try( @a ~~ [1,2,3] );
say "\@a ~~ 3 is ", try( @a ~~ 3 );
say "3 ~~ \@a is ", try( 3 ~~ @a );

say '';

my @b = (4,5,6);
say "\@b is @b";
say "\@b ~~ (4,5,6) is ", try( @b ~~ (4,5,6) );
say "\@b ~~ [4,5,6] is ", try( @b ~~ [4,5,6] );
say "\@b ~~ 3 is ", try( @b ~~ 3 );
say "3 ~~ \@b is ", try( 3 ~~ @b );

say '';
say "\@b ~~ \@a is ", try( @b ~~ @a );

sub try { $_[0] || 0 }

各种情况的输出是你误读文档的线索:

Useless use of a constant (2) in void context at test.pl line 8.
Useless use of a constant (4) in void context at test.pl line 17.
Useless use of a constant (5) in void context at test.pl line 17.
@a is 1 2 3
@a ~~ (1,2,3) is 0
@a ~~ [1,2,3] is 1
@a ~~ 3 is 0
3 ~~ @a is 1

@b is 4 5 6
@b ~~ (4,5,6) is 0
@b ~~ [4,5,6] is 1
@b ~~ 3 is 0
3 ~~ @b is 0

@b ~~ @a is 0