揭秘 Perl glob (*) [英] Demystifying the Perl glob (*)
问题描述
在这个问题中,海报询问了如何在一行中执行以下操作:
sub my_sub {我的 $ref_array = shift;我的@array = @$ref_array;}
根据我对基本 Perl 魔法的了解,我可以通过简单地使用以下内容来避免:
sub my_sub {我的 $ref_array = shift;对于(@$ref_array){#在这里用 $_ 做点什么};#在此处使用 $ref_array->[$element]}
但是在 这个答案中,SO 的一位当地僧侣 tchrist 建议:
sub my_sub {本地 *array = shift();#在这里使用@array}
当我问
<块引用>在尝试学习中级 Perl魔法,我能问一下,你是什么在这里设置什么?你是将@array 的引用设置为已经传入的arrayref?如何你知道你创建了@array 和不是 %array 或 $array?我可以在哪里了解有关此 * 运算符的更多信息(perlop?).谢谢!
有人建议我将其作为新帖子提问,尽管他确实提供了很好的参考.无论如何,来了?有人可以解释什么被分配给什么以及如何创建@array 而不是 %array 或 $array?谢谢.
分配给 glob
*glob = VALUE
包含一些依赖于 VALUE
类型的魔法(即,返回值,比如 Scalar::Util::reftype(VALUE)
).如果VALUE
是对标量、数组、散列或子例程的引用,则仅符号表中的该条目将被覆盖.
这个成语
local *array = shift();#在这里使用@array
当子例程的第一个参数是数组引用时,按文档中的说明工作.如果第一个参数是标量引用,那么只有 $array
而不是 @array
会受到赋值的影响.
一个小演示脚本,看看发生了什么:
无严格;子F{本地*数组=移位;打印 "@array = @array
";打印 "$array = $array
";打印 "\%array = ",%array,"
";打印 "-------------------
";}$array = "原始标量";%array = ("原始" => "哈希");@array = ("原始","数组");$foo = "foo";@foo = ("foo","bar");%foo = ("FOO" => "foo");F ["新建","数组"];# 数组引用F "新标量";# 标量参考F {新" =>哈希"};# 哈希引用F *富;# typeglobF'富';# 不是引用,而是分配变量的名称F '别的东西';# 不是参考F ();# 未定义
<小时>
输出:
<前>@array = 新数组$array = 原始标量%array = 原始哈希------------------@array = 原始数组$array = 新标量%array = 原始哈希------------------@array = 原始数组$array = 原始标量%array = 新哈希------------------@array = foo 栏$数组 = foo%array = FOOfoo------------------@array = foo 栏$数组 = foo%array = FOOfoo------------------@数组 =$数组 =%数组 =------------------@array = 原始数组$array = 原始标量%array = 原始哈希------------------附加文档位于 perlmod
和 perldata
.早在引用成为 Perl 的一部分之前,这个习惯用法有助于将数组和散列传递到子例程中.
In this question the poster asked how to do the following in one line:
sub my_sub {
my $ref_array = shift;
my @array = @$ref_array;
}
which with my knowledge of the basic Perl magic I would avoid by simply using something like:
sub my_sub {
my $ref_array = shift;
for (@$ref_array) {
#do somthing with $_ here
};
#use $ref_array->[$element] here
}
However in this answer one of SO's local monks tchrist suggested:
sub my_sub {
local *array = shift();
#use @array here
}
When I asked
In trying to learn the mid-level Perl magic, can I ask, what is it that you are setting to what here? Are you setting a reference to @array to the arrayref that has been passed in? How do you know that you create @array and not %array or $array? Where can I learn more about this * operator (perlop?). Thanks!
I was suggested to ask it as a new post, though he did give nice references. Anyway, here goes? Can someone please explain what gets assigned to what and how come @array gets created rather than perhaps %array or $array? Thanks.
Assignment to a glob
*glob = VALUE
contains some magic that depends on the type of VALUE
(i.e., return value of, say, Scalar::Util::reftype(VALUE)
). If VALUE
is a reference to a scalar, array, hash, or subroutine, then only that entry in the symbol table will be overwritten.
This idiom
local *array = shift();
#use @array here
works as documented when the first argument to the subroutine is an array reference. If the first argument was instead, say, a scalar reference, then only $array
and not @array
would be affected by the assignment.
A little demo script to see what is going on:
no strict;
sub F {
local *array = shift;
print "@array = @array
";
print "$array = $array
";
print "\%array = ",%array,"
";
print "------------------
";
}
$array = "original scalar";
%array = ("original" => "hash");
@array = ("orignal","array");
$foo = "foo";
@foo = ("foo","bar");
%foo = ("FOO" => "foo");
F ["new","array"]; # array reference
F "new scalar"; # scalar reference
F {"new" => "hash"}; # hash reference
F *foo; # typeglob
F 'foo'; # not a reference, but name of assigned variable
F 'something else'; # not a reference
F (); # undef
Output:
@array = new array $array = original scalar %array = originalhash ------------------ @array = orignal array $array = new scalar %array = originalhash ------------------ @array = orignal array $array = original scalar %array = newhash ------------------ @array = foo bar $array = foo %array = FOOfoo ------------------ @array = foo bar $array = foo %array = FOOfoo ------------------ @array = $array = %array = ------------------ @array = orignal array $array = original scalar %array = originalhash ------------------
Additional doc at perlmod
and perldata
. Back in the days before references were a part of Perl, this idiom was helpful for passing arrays and hashes into subroutines.
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