[问题]请问这段perl代码哪有问题？

enzo1985 · **发表于 :** 2008-02-01 14:23

代码:

#!/usr/bin/perl

my %sue = (
   "name" => "sue",
   age => 45,
);
%john = (
              "name" => "john",
   age => 20,
   );
%peggy = (
   "name" => "peggy",
   age => 16,
   );
#$sue{children} = [\%john, \%peggy];
@son = [\%john, \%peggy];
$sue{children} = \@son;
print "hello  $sue{children}->[1]->{age}";

为什么得不到我想要的值，怎样才能实现打印该值？？谢谢

heejun · **发表于 :** 2008-02-06 2:03

enzo1985 写道:

代码:

#!/usr/bin/perl

my %sue = (
   "name" => "sue",
   age => 45,
);
%john = (
              "name" => "john",
   age => 20,
   );
%peggy = (
   "name" => "peggy",
   age => 16,
   );
#$sue{children} = [\%john, \%peggy];
@son = [\%john, \%peggy];
$sue{children} = \@son;
print "hello  $sue{children}->[1]->{age}";

为什么得不到我想要的值，怎样才能实现打印该值？？谢谢

问题出在这一句

代码:

@son = [\%john, \%peggy];

你在这里多引用了一层,按照你的意思,你想造一个引用数组@son
正确的方法是

代码:

@son = (\%john, \%peggy);

ps:最好加上

代码:

use strict;
use warnings;

heejun · **发表于 :** 2008-02-06 2:07

或者你直接将匿名数组赋值给hash也是可以的

代码:

#!/usr/bin/perl 
use strict;
use warnings;
my %sue = (
   "name" => "sue",
      age => 45,
      );
my %john = (
      "name" => "john",
       age => 20,
        );
my %peggy = (
     "name" => "peggy",
      age => 16,
   );
#$sue{children} = [\%john, \%peggy]; 
$sue{children} =[\%john, \%peggy];
print "hello  $sue{children}->[1]->{age}";

enzo1985 · **发表于 :** 2008-02-07 0:03

的确是那个地方的问题，非常感谢

[问题]请问这段perl代码哪有问题？

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