Java嵌套的通用类型不匹配

Question

在以下示例中：

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 ");
}

为什么对second（b）的调用有效，但对third（a）的调用却无效 ？

我收到以下错误：

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

Answer 1

如果您希望能够fourth使用自List<List<String>>变量进行调用，则需要将签名更改为此：

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

将上面的工作，因为不像List<List<?>>，List<? extends List<?>>是兼容List<List<String>>。这样想：

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

推理很简单。如果你有

private static void fourth(List<List<?>> a) {
    List<?> ohOh = Arrays.asList(new Object());
    a.add(ohOh);
}

然后，如果您 可以这样 调用该方法：

List<List<String>> a = new ArrayList<List<String>>();
fourth(a);
String fail = a.get(0).get(0); // ClassCastException here!