arrays - perl:在使用 "strict refs"时，不能将字符串用作ARRAY引用

Question

我在尝试将数组传递给子项并读取其他值时遇到了这个问题。在此测试脚本中，我传递了数组@players的引用，但传递了读取字符串$ buffer。

use Data::Dumper;

my @players = [];
my @playerscores = [];
my %FORM;

my $buffer = "numplayers=2&changeplayers=3&CHANGEIT=CHANGEIT&player1=a&player2=b&restart=1&newcoords=";

sub testsub {
    my @testee = @$_[0];
    print "in testsub: $testee[0]\n";
}

my @holder = split(/&/,$buffer,);
foreach my $iter (@holder)
{
    my ($name,$value) = split(/=/,$iter);
    $FORM{$name} = $value;
}
$_ = $buffer;
foreach my $key (keys(%FORM))
{
    if($key=~ /player[1-9]/)
    {
        if(!($FORM{$key} eq ""))
        {
            my $holder = $key;
            $holder =~ s/player//;
            $players[$holder] = $FORM{$key};
            $playerscores[$holder] = 0;
            $_ = $buffer;
        }
    }
}

print "\n Data Dumper on player:\n";
print Dumper(@players);
print "\n\n";

print "Check sub:\n";
testsub(\@players,\@playerscores);

输出:

 Data Dumper on player:
$VAR1 = [];
$VAR2 = 'a';
$VAR3 = 'b';


Check sub:
Can't use string ("numplayers=2

&changeplayers=3&CHA"...) as an ARRAY ref while "strict refs" in use at test-str-pool2.pl line 15.

而我期望 print "in testsub: $testee[0]\n";的结果是“a”或“b”。为什么会这样呢？

Answer 1

您错误地声明了数组，这就是Dumper输出为何将[](空数组引用)作为@players中的第一个元素的原因。用:

my @players = ();
my @playerscores = ();

第二个错误来自:

my @testee = @$_[0];

这将尝试取消引用 $_并从结果数组中获取第一个元素。你的意思是:

my @testee = @{$_[0]};

它从 @_中获取第一个元素，然后对其取消引用。