面向对象的Perl构造函数语法 [英] Object-Oriented Perl constructor syntax

问题描述

我对于Perl构造函数中发生了什么感到困惑。我找到了这两个示例 perldoc perlbot 。

I'm a little confused about what is going on in Perl constructors. I found these two examples perldoc perlbot.

package Foo;

#In Perl, the constructor is just a subroutine called new.
sub new {
  #I don't get what this line does at all, but I always see it. Do I need it?
  my $type = shift;

  #I'm turning the array of inputs into a hash, called parameters.
  my %params = @_;

  #I'm making a new hash called $self to store my instance variables?
  my $self = {};

  #I'm adding two values to the instance variables called "High" and "Low".
  #But I'm not sure how $params{'High'} has any meaning, since it was an
  #array, and I turned it into a hash.
  $self->{'High'} = $params{'High'};
  $self->{'Low'} = $params{'Low'};

  #Even though I read the page on [bless][2], I still don't get what it does.
  bless $self, $type;
}

另一个例子是：

package Bar;

sub new {
  my $type = shift;

  #I still don't see how I can just turn an array into a hash and expect things
  #to work out for me.
  my %params = @_;
  my $self = [];

  #Exactly where did params{'Left'} and params{'Right'} come from?
  $self->[0] = $params{'Left'};
  $self->[1] = $params{'Right'};

  #and again with the bless.
  bless $self, $type;
}

这里是使用这些对象的脚本：

And here is the script that uses these objects:

package main;

$a = Foo->new( 'High' => 42, 'Low' => 11 );
print "High=$a->{'High'}\n";
print "Low=$a->{'Low'}\n";

$b = Bar->new( 'Left' => 78, 'Right' => 40 );
print "Left=$b->[0]\n";
print "Right=$b->[1]\n";

我已经把注意到的问题/混淆了。 / p>

I've injected the questions/confusion that I've been having into the code as comments.

推荐答案

为了回答你的问题的主要推力，因为哈希可以初始化为 => value 对，您可以将此列表发送到函数，然后将 @_ 分配给散列。这是在Perl中执行命名参数的标准方法。

To answer the main thrust of your question, since a hash can be initialized as a list of key => value pairs, you can send such a list to a function and then assign @_ to a hash. This is the standard way of doing named parameters in Perl.

例如，

sub foo { 
    my %stuff = @_;
    ...
}

foo( beer => 'good', vodka => 'great' );

这将导致子程序中的％stuff foo 具有两个键的散列， beer 和 vodka ，以及相应的值。

This will result in %stuff in subroutine foo having a hash with two keys, beer and vodka, and the corresponding values.

现在，在OO Perl中，还有一些额外的皱纹。每当你使用箭头（ - > ）运算符来调用一个方法时，箭头左边的任何东西都会停留在 @ 数组。

Now, in OO Perl, there's some additional wrinkles. Whenever you use the arrow (->) operator to call a method, whatever was on the left side of the arrow is stuck onto the beginning of the @_ array.

所以如果你说 Foo-> new（1，2，3） / code>;

So if you say Foo->new( 1, 2, 3 );

然后在你的构造函数中， @_ 看起来像这样：（'Foo'，1，2，3）。

Then inside your constructor, @_ will look like this: ( 'Foo', 1, 2, 3 ).

所以我们使用 shift @ $ 隐含地操作 ，并将其分配给 $ type 。之后， @_ 只剩下我们的名称/值对，为方便起见，我们可以将其直接分配给散列。

So we use shift, which without an argument operates on @_ implicitly, to get that first item out of @_, and assign it to $type. After that, @_ has just our name/value pairs left, and we can assign it directly to a hash for convenience.

然后，对 bless 使用 $ type 所有 bless 做的是引用（在你的第一个例子中一个哈希ref），并说这个引用与特定的包相关联。 Alakazzam，你有一个对象。

We then use that $type value for bless. All bless does is take a reference (in your first example a hash ref) and say "this reference is associated with a particular package." Alakazzam, you have an object.

请记住， $ type 包含字符串'Foo'的我们的包。如果您不为 bless 指定第二个参数，它将使用当前包的名称，这在本示例中也可以使用，但是将不继承的构造函数。

Remember that $type contains the string 'Foo', which is the name of our package. If you don't specify a second argument to bless, it will use the name of the current package, which will also work in this example but will not work for inherited constructors.

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