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。
此解視Semaphore為鎖，以保證同一時刻單線程的順序執行。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，分別調用R1,R2,R3的方法多次，控制臺輸出對應各方法名拼接執行該方法的線程名。多次執行結果各不相同：
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
方法能調用，多線程下卻無法保證方法的順序執行。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);
進入R2、R3方法的線程會執行acquire()方法，而B、C中的計數器為0獲取不到許可，阻塞直到一個可用，或者線程被中斷，不能繼續執行。R1方法中A尚有1個許可可拿到，方法執行，并給B發布一個許可，若B先于A執行acquire()，此時B為阻塞狀態，則獲取到剛剛發布的許可，該線程被重新啟用。
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