我不明白为什么在以下情况下Java编译器会向我发出“未经检查的转换”警告:

我有这个课:

public class NodeTree<T> {
   T value;
   NodeTree parent;
   List<NodeTree<T>> childs;

   NodeTree(T value, NodeTree parent) {
       this.value = value;
       this.parent = parent;
       this.childs = null;
   }

   public T getValue() { return value; }
   public void setValue(T value) { this.value = value; }

   public NodeTree getParent() { return parent; }
   public void setParent(NodeTree parent) { this.parent = parent; }

   public List<NodeTree<T>> getChilds() {
       if (this.childs == null) {
           this.childs = new LinkedList<NodeTree<T>>();
       }
       return this.childs;
   }
}


在主要课程中,我有以下说明:

NodeTree node = new NodeTree<Integer>(10, null);

NodeTree<Integer> child = new NodeTree<Integer>(20, node);
List<NodeTree<Integer>> childs = node.getChilds();

childs.add(child);


我无法解释为什么我在这种类型的getChilds()行上得到警告:

warning: [unchecked] unchecked conversion
List<NodeTree<Integer>> childs = node.getChilds();
                                               ^
required: List<NodeTree<Integer>>
found:    List
1 warning


getChilds()函数不返回List类型,它返回List >类型。

请帮助我理解。

最佳答案

编写NodeTree<Integer> node = new NodeTree<>(10, null);会更好吗
代替NodeTree node = new NodeTree<Integer>(10, null);?然后,编译器将知道node的类型参数。

09-30 14:15