Groovy散点算子 [英] Groovy spread-dot operator

问题描述

我不确定何时需要使用扩散点运算符。如果我有一个列表，例如：

  def animals = ['ant'，'buffalo'，'canary'，'dog '] 
  

因为 animals 和成员的列表有一个size（）方法，那么为了让列表元素（而不是列表本身）被调用，我需要使用：

< pre $ animals * .size（）


而非：

  animals.size（）
  

但是，请考虑以下表达式，它将返回String类中可用的方法名称列表

  String.metaClass.methods.name 
  

Groovy如何知道（不使用扩展点运算符）我希望从 String.metaClass.methods 返回的列表的每个元素中检索名称属性，而不是名单本身？在这种情况下是否有一组规则可供使用，如：

• 如果List上存在method / property，则在List li>
  
• 否则，尝试为List的每个元素调用它
  
  $ b

  谢谢，
  Don

  解决方案

  简短的答案是 DefaultGroovyMethods为所有集合添加一个getAt方法，它循环访问Collection并收集属性值每。

  
  
  

  如果您对长答案感兴趣，我写了一个博文在前一段时间里，它在这个确切的话题上潜入metaClass兔子洞中。

  
  

  I'm unsure about when exactly I need to use the spread-dot operator. If I have a list such as:

  def animals = ['ant', 'buffalo', 'canary', 'dog']
  

  Because both animals and the members of the list have a size() method, then in order to have the method invoked on the elements of the list (rather than the list itself), I need to use:

  animals*.size()
  

  rather than:

  animals.size()
  

  However, consider the following expression which returns a list of the method names that are available on the String class

  String.metaClass.methods.name
  

  How does Groovy know (without using the spread-dot operator) that I want the name property to be retrieved from each element of the list returned by String.metaClass.methods, rather than the list itself? Is there a set of rules to use in this case such as:

  • If method/property exists on List, invoke it on List
  • Otherwise try to invoke it for each element of List

  Thanks, Don

  解决方案

  The short answer is that DefaultGroovyMethods adds a "getAt" method to all Collections that iterates through the Collection and collects the property value for each.

  If you're interested in the long answer, I wrote up a blog post that dives down the metaClass rabbit hole on this exact topic a while ago.

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