使用CAS代替synchronized

r囧r小猫 2023-10-01 18:16 21阅读 0赞

在开发当中需要经常用到synchronized保证代码[线程安全][Link 1]，在竞争条件下会阻塞等待资源，如果允许竞争不到资源返回失败，就可以使用cas减少阻塞时间。先来看一个cas的单例模式。

public class NonBlock {
        
     
    	private static volatile NonBlock nonBlock;
     
    	private static AtomicBoolean atomicBoolean = new AtomicBoolean(false);
     
    	public static NonBlock getInstance() {
        
    		if (nonBlock == null) {
        
    			if (atomicBoolean.compareAndSet(false, true)) {
        
    				nonBlock = new NonBlock();
    			}
    		}
    		return nonBlock;
    	}
    }

在这个单例模式中，不同于synchronized的阻塞，多线程环境下，getInstance确保只会创建一个对象的情况下，可能返回的nonBlock是一个空对象。但，如果允许返回空对象的情况下，

public class RedPacket {
        
     
    		private long balance;
     
    		private int num;
     
    		public RedPacket(long balance, int num) {
        
    			this.balance = balance;
    			this.num = num;
    		}
     
    		public long get() {
        
    			if (balance < 1 || num < 1) {
        
    				return -1;
    			}
    			if (num == 1) {
        
    				long result = balance;
    				balance = 0;
    				num--;
    				return result;
    			}
    			long average = balance / num;
    			long result = ThreadLocalRandom.current().nextLong(1, average * 2);
    			balance -= result;
    			num--;
    			return result;
    		}

我们还可以使用cas达到非阻塞的目的，这样能保证线程安全，出现竞争情况就提示抢失败，确点就是提示抢失败还可能余额大于0，先来不一定能抢到，后来人还能抢。

public class RedPacket {
        
     
    		private long balance;
     
    		private AtomicInteger num;
     
    		public RedPacket(long balance, int num) {
        
    			this.balance = balance;
    			this.num = new AtomicInteger(num);
    		}
     
    		public long get() {
        
    			int number = num.get();
    			long balan = balance;
    			if (balan < 1 || number < 1) {
        
    				return -1;
    			}
    			if (num.compareAndSet(number, number - 1)) {
        
    				if (number - 1 == 0) {
        
    					balance = 0;
    					return balan;
    				}
    				long average = balan / number;
    				long result = ThreadLocalRandom.current().nextLong(1, average * 2);
    				balance -= result;
    			}
    			return -1;
    		}
    	}

[Link 1]: https://so.csdn.net/so/search?q=%E7%BA%BF%E7%A8%8B%E5%AE%89%E5%85%A8&spm=1001.2101.3001.7020