Java 中的 ConcurrentSkipListSet removeAll()方法
原文:https://www . geeksforgeeks . org/concurrentskiplistset-removeall-method-in-Java/
java . util . concurrentskiplistset的 removeAll() 方法是 Java 中的一个内置函数，它从这个集合中返回并移除指定集合中包含的所有元素。如果指定的集合也是集合，则此操作会有效地修改该集合，使其值为两个集合的不对称集合差。
语法:
public boolean removeAll(Collection c)
参数:该函数接受单个参数 c
返回值:如果该集合因调用而改变，则函数返回真。
异常:函数抛出以下异常:
-
class castexception–如果该集合中一个或多个元素的类型与指定集合不兼容 NullPointerException* – if the specified collection or any of its elements are null
下面的程序说明了 concurrentskiplistset . removeall()方法:
程序 1:
```java // Java program to demonstrate removeAll() // method of ConcurrentSkipListSet
import java.util.concurrent.*; import java.util.ArrayList; import java.util.List;
class ConcurrentSkipListSetremoveAllExample1 { public static void main(String[] args) {
// Initializing the List List list = new ArrayList();
// Adding elements in the list for (int i = 1; i <= 10; i += 2) list.add(i);
// Contents of the list System.out.println("Contents of the list: " + list);
// Initializing the set ConcurrentSkipListSet set = new ConcurrentSkipListSet();
// Adding elements in the set for (int i = 1; i <= 10; i++) set.add(i);
// Contents of the set System.out.println("Contents of the set: " + set);
// Remove all elements from the set which are in the list set.removeAll(list);
// Contents of the set after removal System.out.println("Contents of the set after removal: " + set); } } ```
Output:
```java Contents of the list: [1, 3, 5, 7, 9] Contents of the set: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] Contents of the set after removal: [2, 4, 6, 8, 10]
```
程序 2: 在 removeAll()中显示 NullPOinterException 的程序。
```java // Java program to demonstrate removeAll() // method of ConcurrentSkipListSet
import java.util.concurrent.*; import java.util.ArrayList; import java.util.List;
class ConcurrentSkipListSetremoveAllExample1 { public static void main(String[] args) {
// Initializing the set ConcurrentSkipListSet set = new ConcurrentSkipListSet();
// Adding elements in the set for (int i = 1; i <= 10; i++) set.add(i);
// Contents of the set System.out.println("Contents of the set: " + set);
try { // Remove all elements from the set which are null set.removeAll(null); } catch (Exception e) { System.out.println("Exception: " + e); } } } ```
Output:
```java Contents of the set: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] Exception: java.lang.NullPointerException
```
参考:
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