Java泛型 - 真正的无界通配符是什么？

问题描述

如果我有以下代码：

  public static void main（String [] args）{
 List <整数> l2 = new ArrayList< Integer>（）; 
列表<？> l3 = l2; 
 test（l2）; 
 test（l3）; 
 
 
 
 public static void test（List<> l）{
 if（l instanceof List<>）
 System .out.println（ 真）; 
} 
  

这将打印：

  true 
 true 
  

理解，<？> 是一个可定义的类型，这意味着它有一些捕获类型（不管那种类型是什么），它在运行时可用。

问题：

a。在测试方法中，它是否知道l2具有整数类型（因为它在方法调用之前已被擦除）？它是如何翻译的，以便l（来自l2）是instanceof List ？

b。那么L3呢？我不相信<？>

code>被通用化。这只是在不使用原始表单（ List ）的情况下引用generified类型的唯一方法。在这两种情况下，您只需执行完全相同的操作即可：

编辑

的确，我刚刚验证过无论您在中使用 List > 或 List instanceof 。

If i have the following code :

public static void main(String [] args) {
        List <Integer> l2 = new ArrayList <Integer>();
        List <?> l3 = l2;
        test(l2);
        test(l3);
}


public static void test(List <?> l) {
        if (l instanceof List<?>)
            System.out.println("true");
}

This will print:

true
true

From what i understand, <?> is a reifiable type, which means it has some capture type (whatever that type is) which is available at run time.

Questions:
a. In test method, does it know that l2 has integer type (since it has been erased prior to the method call)? How does it translate so that l (from l2) is instanceof List <?>?
b. What about l3? How does it translate that?

解决方案

I don't believe the <?> is reified. It is simply the only way to refer to a generified type without using the raw form (List). In both cases you are simply doing the exact same operation as:

if (l instanceof List)
   ...

Edit

Indeed I have just verified that they generate absolutely identical bytecode whether you use List<?> or List in the instanceof.

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