Java ConcurrentSkipListSet removeAll()方法

java.util.concurrent.ConcurrentSkipListSet 的 removeAll() 方法是Java中的一个内置函数，它返回从这个集合中删除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，这个操作可以有效地修改这个集合，使其值为两个集合的不对称集合差。

语法

public boolean removeAll(Collection c)

参数： 该函数接受一个单一的参数c

返回值： 如果该集合因调用而发生变化，则该函数返回真。

异常： 该函数会抛出下列异常。

• ClassCastException - 如果这个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。

程序1 :

// 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<Integer> list = new ArrayList<Integer>();
  
        // 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<Integer>
            set = new ConcurrentSkipListSet<Integer>();
  
        // 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<Integer>
            set = new ConcurrentSkipListSet<Integer>();
  
        // 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

参考资料

https://docs.oracle.com/javase/8/docs/api/java/util/concurrent/ConcurrentSkipListSet.html#removeAll-java.util.Collection-