Java中的 ConcurrentSkipListSet removeAll() 方法

Java .util.concurrent.ConcurrentSkipListSet的removeAll()方法是Java中的一个内置函数，它返回从该集合中删除的所有包含在指定集合中的元素。如果指定的集合也是一个集合，这个操作有效地修改了这个集合，使得它的值是两个集合的不对称集合差。

句法：

public boolean removeAll(Collection c)

参数：函数接受单个参数c

返回值：如果此集合因调用而更改，则该函数返回 true。

异常：该函数抛出以下异常：

ClassCastException – 如果此集合中的一个或多个元素的类型与指定的集合不兼容

NullPointerException – 如果指定的集合或其任何元素为空

下面的程序说明了 ConcurrentSkipListSet.removeAll() 方法：

方案一：

// 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);
    }
}

输出：

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 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);
        }
    }
}

输出：

Contents of the set: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Exception: java.lang.NullPointerException

参考：

https://docs.oracle.com/javase/8/docs/api/java Java