Java ConcurrentSkipListSet removeAll()用法及代碼示例

java.util.concurrent.ConcurrentSkipListSet的removeAll()方法是Java中的一個內置函數，該函數從該集合中移除指定集合中包含的所有元素。如果指定的集合也是一個集合，則此操作會有效地修改此集合，以使其值為兩個集合的非對稱集合差異。

用法:

public boolean removeAll(Collection> c)

參數：該函數接受單個參數c

返回值：如果此設置由於調用而改變，則函數返回true。

異常：該函數引發以下異常：

ClassCastException–如果該集合中一個或多個元素的類型與指定的集合不兼容
NullPointerException –如果指定的集合或其任何元素為空

以下示例程序旨在說明ConcurrentSkipListSet.removeAll()方法：

示例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); 
    } 
}

輸出：

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]

示例2:在removeAll()中顯示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); 
        } 
    } 
}

輸出：

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

參考：

相關用法

注：本文由純淨天空篩選整理自rupesh_rao大神的英文原創作品 ConcurrentSkipListSet removeAll() method in Java。非經特殊聲明，原始代碼版權歸原作者所有，本譯文未經允許或授權，請勿轉載或複製。