perl – 什么是’sub bar {{_ _ [1] => $_ [2]}}`完全可以吗

发布时间：2020-12-15 21:31:06 所属栏目：大数据来源：网络整理

导读：我不明白这个例子中的糖迹象在某个地方, sub bar { +{$_[1] = $_[2]} } 我写了这个,我没有看到任何差异： use Data::Dumper;# Not any differences heremy $foo = {value = 55};my $bar = +{value = 55};print Dumper $foo;print Dumper $bar;# Oh ! Here th

我不明白这个例子中的糖迹象在某个地方,

sub bar { +{$_[1] => $_[2]} }

我写了这个,我没有看到任何差异：

use Data::Dumper;

# Not any differences here
my $foo =  {value => 55};
my $bar = +{value => 55};

print Dumper $foo;
print Dumper $bar;

# Oh ! Here there is something...
sub foo {  {$_[1] => $_[2]} };
sub bar { +{$_[1] => $_[2]} };

print Dumper foo('value',55);    
print Dumper bar('value',55);

foo返回

$VAR1 = 55;
$VAR2 = undef;

酒吧回来

$VAR1 = {
          '55' => undef
        };

解决方法

它有助于解析器区分匿名哈希和代码块.

引用Learning Perl Objects,References & Modules

because blocks and anonymous hash constructors both use curly braces in roughly the same places in the syntax tree,the compiler has to make ad hoc determinations about which of the two you mean. If the compiler ever decides incorrectly,you might need to provide a hint to get what you want. To show the compiler that you want an anonymous hash constructor,put a plus sign before the opening curly brace: +{ … }. To be sure to get a block of code,just put a semicolon (representing an empty statement) at the beginning of the block: {; … }.

或从地图上的文档功能：

"{" starts both hash references and blocks,so "map { ..." could
be either the start of map BLOCK LIST or map EXPR,LIST. Because
Perl doesn't look ahead for the closing "}" it has to take a guess
at which it's dealing with based on what it finds just after the
"{". Usually it gets it right,but if it doesn't it won't realize
something is wrong until it gets to the "}" and encounters the
missing (or unexpected) comma. The syntax error will be reported
close to the "}",but you'll need to change something near the "{"
such as using a unary "+" or semicolon to give Perl some help:

    %hash = map {  "L$_" => 1  } @array # perl guesses EXPR. wrong
    %hash = map { +"L$_" => 1  } @array # perl guesses BLOCK. right
    %hash = map {; "L$_" => 1  } @array # this also works
    %hash = map { ("L$_" => 1) } @array # as does this
    %hash = map {  lc($_) => 1  } @array # and this.
    %hash = map +( lc($_) => 1 ),@array # this is EXPR and works!

    %hash = map  ( lc($_),1 ),@array # evaluates to (1,@array)

or to force an anon hash constructor use "+{":

    @hashes = map +{ lc($_) => 1 },@array # EXPR,so needs
                                           # comma at end

to get a list of anonymous hashes each with only one entry apiece.

（编辑：李大同）

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perl – 什么是’sub bar {{_ _ [1] =&gt; $_ [2]}}`完全可以吗

解决方法

perl – 什么是’sub bar {{_ _ [1] => $_ [2]}}`完全可以吗