ArrayList <？超级号码&gt;和双

问题描述

通过：
$ b

那么为什么

   Number 或超类型 - 它是一个子类。你可以这样写： 
 
 
  //有效的
 ArrayList<扩展Number> psupn1 = new ArrayList< Integer>（）; 
  
 ...因为这是相反的。在这一点上，你可以写： 
 
 
  Number x = psupn1.get（0）; 
  
因为列表中的任何元素都保证可以转换到 数。这是所有转换所需的 -  至泛型类型参数或 。
 
From http://www.angelikalanger.com/GenericsFAQ/FAQSections/TypeArguments.html#FAQ103:
So why
ArrayList<? super Number> psupn1 = new ArrayList<Number>();
psupn1.add(new Double(2));
compiled?
Double is not supertype of Number but subclass of Number...
Edit 1:
    ArrayList<? super Number> pextn1 = new ArrayList<Number>();
    psupn1.add(new Integer(2));
    psupn1.add(new Double(2));
    psupn1.add(new Float(2));
    for(Number n : psupn1){ // [Invalid] Number should be change to
    // Object even if I can only add subtype of Number??

    }
 解决方案 
You can add a Double to that, because whatever the type parameter E is, it's guaranteed to be either Number or a supertype... which means you can definitely convert from Double to E. You wouldn't be able to do:
Number x = psupn1.get(0);
though.
Think about it, and try to create lists which would logically break this. For example, you can't use:
// Invalid
ArrayList<? super Number> psupn1 = new ArrayList<Integer>();
psupn1.add(new Double(2));
because Integer isn't either Number or a supertype - it's a subclass. You can write:
// Valid
ArrayList<? extends Number> psupn1 = new ArrayList<Integer>();
... because that's the other way round. At that point you can write:
Number x = psupn1.get(0);
because any element in the list is guaranteed to be convertible to Number. It's all about which way the conversions are required - to the generic type parameter or from it.
                        
这篇关于ArrayList <？超级号码&gt;和双的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！