Joa*_*him 0 tree perl data-structures
我将在周末学习Perl的面试.为了更深入地理解我正在尝试实现树类.
#use strict;
#use warnings;
package Tree;
sub new {
my $class = shift @_;
my $content = shift @_;
my @array = shift @_;
return bless { "content" => $content, "array" => @array }, $class;
}
sub num_children {
my $self = shift @_;
my @array = $self->{"array"};
return scalar @array;
}
return 1;
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为了测试(错误的)树类,我实现了以下测试脚本.
#!/usr/bin/perl
require Tree;
my $t = Tree->new("#", undef);
my $tt = Tree->new("*", undef);
my $tttt = Tree->new("-", undef);
my $ttttt = Tree->new(".", undef);
my @list = ();
push @list, $tt;
push @list, $t;
push @list, $tttt;
push @list, $ttttt;
my $ttt = Tree->new("+", @list);
print $ttt->num_children();
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不幸的是,输出1不是我的预期4.我假设数组以某种方式被切断或不自觉地转换为标量.有任何想法吗?
主要问题是您无法将数组作为单个值传递 - 您必须传递引用.
此外,你永远不应该评论use strict和use warnings.它们是有价值的调试工具,如果您在启用它们时收到错误消息,则应该修复它们正在标记的错误.
这是一个工作 Tree.pm
use strict;
use warnings;
package Tree;
sub new {
my $class = shift;
my ($content, $array) = @_;
return bless { content => $content, array => $array }, $class;
}
sub num_children {
my $self = shift;
my $array = $self->{array};
return scalar @$array;
}
1;
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和调用程序tree_test.pl.请注意,您应该use而不是require模块.
#!/usr/bin/perl
use strict;
use warnings;
use Tree;
my @list = map { Tree->new($_) } ('#', '*', '-', '.');
my $ttt = Tree->new('+', \@list);
print $ttt->num_children, "\n";
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产量
4
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