Java嵌套的泛型类型不匹配 [英] Java nested generic type mismatch

问题描述

在下面的例子中：

  public static void main（String [] args）{
 
列表< String> b = new ArrayList< String>（）; 
先（b）; $ b b秒（b）; 
 
列表< List< String>> a = new ArrayList< List< String>>（）; 
第三（a）; 
第四（a）; //不工作
 
} 
 
私人静态< T> void（first< T> a）{
 System.out.println（List of T）; 
 
 
 private static void second（List<？> a）{
 System.out.println（List of anything）; 
} 
 
私人静态< T> void third（List< List< T>> a）{
 System.out.println（List of T of List）; 
 
 
 private static void fourth（List< List<>> a）{
 System.out.println（任何列表的列表）; 
 
 
 
 
 
为什么第二个（b）的调用有效，但是致电第四（a）不？
 
 
 
我收到以下错误： 
 
 
类型TestTest中的方法fourth（List< List<>）不适用于参数（`List< List< String>>`）
  
 
 
解决方案
如果您希望能够调用第四与一个 List< List< String>> 参数，那么您需要将签名更改为：
  private static void fourth（List< ;? extends List<>> a）{
 System.out.println（List任何事物的清单）; 
} 
  
因为不像 List< List< ？>> ，列表< ;? extends List<>> 与 List< List< String>> 兼容。以这种方式考虑： 
 
 
  List< List< String>> original = null; 
列表< ;?扩展List<>>>好的=原创; //这工作
列表<？> ok2 =原创; //这个
 List< List<>>> notOk = original; //这不是
 
 List< Integer> original = null; 
列表< ;?扩展Number>好的=原创; //这工作
列表<？> ok2 =原创; //这个
列表< Number> notOk = original; //这不是
  
推理很简单。如果您有 
 
 
  private static void fourth（List< List<>> a）{
 List< ？> ohOh = Arrays.asList（new Object（））; 
 a.add（ohOh）; 
} 
  
然后，如果可以将该方法称为例如： 
 
 
  List< List< String>> a = new ArrayList< List< String>>（）; 
第四（a）; 
 String fail = a.get（0）.get（0）; // ClassCastException here！ 
  
 
In the following example:
public static void main(String[] args) {

    List<String> b = new ArrayList<String>();
    first(b);
    second(b);

    List<List<String>> a = new ArrayList<List<String>>();
    third(a);
    fourth(a);  // doesnt work

}

private static <T> void first(List<T> a){
    System.out.println("List of T");
}

private static void second(List<?> a){
    System.out.println("List of anything ");
}

private static <T> void third(List<List<T>> a){
    System.out.println("List of a List of T ");
}

private static void fourth(List<List<?>> a){
    System.out.println("List of a List of anything ");
}
Why does the call to second(b) work, but the call to fourth(a) doesn't ?


I get the following error:
The method fourth(List<List<?>>) in the type `TestTest` is not applicable for the arguments (`List<List<String>>`)

解决方案
If you want to be able to call fourth with a List<List<String>> argument, then you'll need to change your signature to this:
private static void fourth(List<? extends List<?>> a){
    System.out.println("List of a List of anything ");
}
The above will work because unlike List<List<?>>, List<? extends List<?>> is compatible with List<List<String>>. Think of it this way:
List<List<String>> original = null;
List<? extends List<?>> ok  = original; // This works
List<?> ok2                 = original; // So does this
List<List<?>> notOk         = original; // This doesn't

List<Integer> original      = null;
List<? extends Number> ok   = original; // This works
List<?> ok2                 = original; // So does this
List<Number> notOk          = original; // This doesn't
The reasoning is simple. If you had
private static void fourth(List<List<?>> a) {
    List<?> ohOh = Arrays.asList(new Object());
    a.add(ohOh);
}
And then if you could call that method as such:
List<List<String>> a = new ArrayList<List<String>>();
fourth(a);
String fail = a.get(0).get(0); // ClassCastException here!


                        
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