Java信號量Semaphore的實現(xiàn)

近日于LeetCode看題遇1114 按序打印，獲悉一解法使用了Semaphore，順勢研究，記心得于此。

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此解視Semaphore為鎖，以保證同一時刻單線程的順序執(zhí)行。在此原題上，我作出如下更改。

package test;
 
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Semaphore;
 
public class SemaphoreDemo {
 static Semaphore A;
 static Semaphore B;
 static Semaphore C;
 
 public static void main(String[] args) throws InterruptedException { 
    A = new Semaphore(1);
    B = new Semaphore(0);
    C = new Semaphore(0);
    
    ExecutorService ex=Executors.newFixedThreadPool(10);
    
 for (int i = 0; i <7; i++) {
  ex.execute(new R1());
  ex.execute(new R2());
  ex.execute(new R3());
 }
    ex.shutdown();
  }   
 
 public static class R1 implements Runnable{
 @Override
 public void run() {
  try {
//  A.acquire();
  System.out.println("1"+Thread.currentThread().getName());
//  B.release();
  } catch (Exception e) {
  e.printStackTrace();
  }
 } 
  }
 
 public static class R2 implements Runnable{
 @Override
 public void run() {
  try {
//  B.acquire();
  System.out.println("2"+Thread.currentThread().getName());
//  C.release();
  } catch (Exception e) {
  e.printStackTrace();
  }
 } 
  }
 
 public static class R3 implements Runnable{
 @Override
 public void run() {
  try {
//  C.acquire();
  System.out.println("3"+Thread.currentThread().getName());
//  A.release();
  } catch (Exception e) {
  e.printStackTrace();
  }
 } 
  }
 
}

10個線程的常量池中，分別調(diào)用R1,R2,R3的方法多次，控制臺輸出對應(yīng)各方法名拼接執(zhí)行該方法的線程名。多次執(zhí)行結(jié)果各不相同：

1pool-1-thread-1
2pool-1-thread-2
1pool-1-thread-4
3pool-1-thread-6
2pool-1-thread-5
3pool-1-thread-3
1pool-1-thread-7
2pool-1-thread-8
3pool-1-thread-9
3pool-1-thread-1
2pool-1-thread-8
1pool-1-thread-4
3pool-1-thread-1
1pool-1-thread-2
2pool-1-thread-9
1pool-1-thread-10
3pool-1-thread-1
2pool-1-thread-5
1pool-1-thread-6
3pool-1-thread-4
2pool-1-thread-8

1pool-1-thread-1
2pool-1-thread-2
3pool-1-thread-3
1pool-1-thread-4
2pool-1-thread-5
3pool-1-thread-6
1pool-1-thread-7
2pool-1-thread-8
3pool-1-thread-9
1pool-1-thread-10
3pool-1-thread-1
1pool-1-thread-4
2pool-1-thread-8
3pool-1-thread-3
2pool-1-thread-10
1pool-1-thread-2
2pool-1-thread-9
3pool-1-thread-4
1pool-1-thread-7
3pool-1-thread-6
2pool-1-thread-5

方法能調(diào)用，多線程下卻無法保證方法的順序執(zhí)行。使用Semaphore后，代碼為：

package test;
 
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Semaphore;
 
public class SemaphoreDemo {
 static Semaphore A;
 static Semaphore B;
 static Semaphore C;
 
 public static void main(String[] args) throws InterruptedException { 
    A = new Semaphore(1);
    B = new Semaphore(0);
    C = new Semaphore(0);
    
    ExecutorService ex=Executors.newFixedThreadPool(10);
    
 for (int i = 0; i <7; i++) {
  ex.execute(new R1());
  ex.execute(new R2());
  ex.execute(new R3());
 }
    ex.shutdown();
  }   
 
 public static class R1 implements Runnable{
 @Override
 public void run() {
  try {
  A.acquire();
  System.out.println("1"+Thread.currentThread().getName());
  B.release();
  } catch (Exception e) {
  e.printStackTrace();
  }
 } 
  }
 
 public static class R2 implements Runnable{
 @Override
 public void run() {
  try {
  B.acquire();
  System.out.println("2"+Thread.currentThread().getName());
  C.release();
  } catch (Exception e) {
  e.printStackTrace();
  }
 } 
  }
 
 public static class R3 implements Runnable{
 @Override
 public void run() {
  try {
  C.acquire();
  System.out.println("3"+Thread.currentThread().getName());
  A.release();
  } catch (Exception e) {
  e.printStackTrace();
  }
 } 
  }
 
}

多次運行結(jié)果皆能保證1、2、3的順序：

1pool-1-thread-1
2pool-1-thread-2
3pool-1-thread-3
1pool-1-thread-4
2pool-1-thread-5
3pool-1-thread-6
1pool-1-thread-7
2pool-1-thread-8
3pool-1-thread-9
1pool-1-thread-10
2pool-1-thread-1
3pool-1-thread-2
1pool-1-thread-3
2pool-1-thread-4
3pool-1-thread-5
1pool-1-thread-6
2pool-1-thread-9
3pool-1-thread-7
1pool-1-thread-10
2pool-1-thread-8
3pool-1-thread-1

附上api文檔鏈接 Semaphore

A = new Semaphore(1);
B = new Semaphore(0);
C = new Semaphore(0);

進(jìn)入R2、R3方法的線程會執(zhí)行acquire()方法，而B、C中的計數(shù)器為0獲取不到許可，阻塞直到一個可用，或者線程被中斷，不能繼續(xù)執(zhí)行。R1方法中A尚有1個許可可拿到，方法執(zhí)行，并給B發(fā)布一個許可，若B先于A執(zhí)行acquire()，此時B為阻塞狀態(tài)，則獲取到剛剛發(fā)布的許可，該線程被重新啟用。

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