Perl指定pos触发副本吗？

我正在写一个小的lexing实用程序.经常出现的一件事是搜索从给定偏移开始的模式…并返回匹配的字符串(如果有的话).

为了支持反复尝试使用令牌,我需要我的函数将pos设置为在匹配之后如果我们成功并且在我们开始搜索的地方(如果我们不是).

例如

my $string = "abc";
consume($string,qr/b/,1);
printf "%sn",pos($string); # should print 2

pos($string) = 0; # reset the pos,just to demonstrate
                  # the intended behavior when there isn't a match

consume($string,qr/z/,pos($string); # should print 1

这是一个返回正确的东西但没有正确设置pos的实现.

package TokenConsume;
use strict;
use warnings;

use Exporter qw[import];
our @EXPORT_OK = qw[consume];

sub consume {
    my ($str,$pat,$pos) = @_;
    pos($str) = $pos;
    my $out = undef;
    if ($str =~ $pat) {
        $out = substr $str,$-[0],($+[0] - $-[0]);
        pos($str) = $+[0];
    } else {
        pos($str) = $pos;
    }
    return $out;
}

这是模块测试套件的示例测试

do {
    my $str = "abc";
    pos($str) = 0;
    my $res = consume($str,1);
    is($res,undef,"non-first: failed match should capture nothing");
    is(pos($str),1,"non-first: failed match should return pos to beginning of search");
};

它失败并显示以下消息(另一个测试失败)：

#   Failed test 'non-first: failed match should return pos to beginning of search'
#   at t/test_tokenconsume.t line 38.
#          got: '0'
#     expected: '1'
# Looks like you failed 2 tests of 7.

我可以通过传入字符串引用并稍微更改API来解决此问题.这是完整性的新实现.

sub consume {
    my ($str_ref,$pos) = @_;
    pos($$str_ref) = $pos;
    my $out = undef;
    if ($$str_ref =~ $pat) {
        $out = substr $$str_ref,($+[0] - $-[0]);
        pos($$str_ref) = $+[0];
    } else {
        pos($$str_ref) = $pos;
    }
    return $out;
}

那么,这里发生了什么？除非我使用引用,为什么pos(…)的赋值不会传播回原始值？