Question

我在同一个哈希上创建子程序外部和内部的键。但是，在子例程之后，调用子例程之前创建的键中的值现在被解释为数组引用。

#!/usr/bin/perl
use module;
use strict;
use warnings;
my %hash;
my $count = 0;

my @array = ("a", "b", "c", "d");

for my $letter (@array) {
    $hash{$letter} = $count;
    $count++;
}

# need "\" to pass in hash otherwise changes
# will get lost outside of subroutine
foreach my $x (sort keys %hash) {
    print "first $hash{$x}\n";
}

module::add_ten(\%hash);

foreach my $p (sort keys %hash) {
    # $hash{$p} is printing array references, but before it was
    # printing the value I desired.  What did the subroutine do?
    print "second $hash{$p} $hash{$p}->{ten}\n";
}

这是带有子程序的模块

package module;

sub add_ten {
my $count = 10;
    # this passes the full array as reference
    my ($hash_ref) = @_; # $hash_ref is actually %hash (yes, the % is not a typo)
    my @keys = keys $hash_ref;
    foreach my $ltr (sort keys $hash_ref) {
        $hash_ref->{$ltr} = { ten => $count };
        $count++;
    }
}
1;

这是输出：

first 0
first 1
first 2
first 3
second HASH(0x7ff0c3049c50) 10
second HASH(0x7ff0c3049bc0) 11
second HASH(0x7ff0c3049b90) 12
second HASH(0x7ff0c3049b60) 13

我期待输出为：

first 0
first 1
first 2
first 3
second 0 10
second 1 11
second 2 12
second 3 13

我修改了我的模块：

package module;

sub add_ten {
    my $count = 10;
    # this passes the full array as reference
    my ($hash_ref) = @_; # $hash_ref is actually %hash (yes, the % is not a typo)
    my @keys = keys $hash_ref;
    foreach my $ltr (sort keys $hash_ref) {
        $hash_ref->{$ltr}{ten}=$count;
        $count++;
    }
}
1;

和主脚本（需要注释use strict以使其工作）：

#!/usr/bin/perl
use module;
#use strict;
use warnings;
my %hash;
my $count = 0;

my @array = ("a", "b", "c", "d");

for my $letter (@array) {
    $hash{$letter} = $count;
    $count++;
}

# need "\" to pass in hash otherwise changes
# will get lost outside of subroutine
foreach my $x (sort keys %hash) {
    print "first $hash{$x}\n";
}

module::add_ten(\%hash);

foreach my $p (sort keys %hash) {
    print "second $hash{$p} $hash{$p}{ten}\n";
}

但这是我想要达到的目的。

Answer 1

$hash_ref是对%hash的引用，因此当您更改$hash_ref引用的哈希元素的值时，您将更改哈希值{{1 }}

这意味着当你做

时

%hash

你在做什么

 $hash_ref->{$ltr} = { ten => $count };

$hash{a} = { ten => 10 };不再包含零并不奇怪。您必须更改数据结构。您可以使用以下内容：

$hash{a}

在perl

1 个答案: