From e4643b907fbf394a0cfbdc7b093949d2a04bb436 Mon Sep 17 00:00:00 2001
From: Hui <zonghui_zhou@foxmail.com>
Date: Sun, 10 May 2020 00:08:05 +0800
Subject: [PATCH] =?UTF-8?q?=E8=A1=A5=E5=85=85=E4=BD=BF=E7=94=A8Semaphore?=
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---
 .../A1B2C3/Print1A2B3C_Semaphore.java         | 52 +++++++++++++++++++
 1 file changed, 52 insertions(+)
 create mode 100644 src/main/java/com/mashibing/juc/c_026_00_interview/A1B2C3/Print1A2B3C_Semaphore.java

diff --git a/src/main/java/com/mashibing/juc/c_026_00_interview/A1B2C3/Print1A2B3C_Semaphore.java b/src/main/java/com/mashibing/juc/c_026_00_interview/A1B2C3/Print1A2B3C_Semaphore.java
new file mode 100644
index 0000000..4f84f65
--- /dev/null
+++ b/src/main/java/com/mashibing/juc/c_026_00_interview/A1B2C3/Print1A2B3C_Semaphore.java
@@ -0,0 +1,52 @@
+package com.mashibing.juc.c_026_00_interview.A1B2C3;
+
+import java.util.concurrent.Semaphore;
+
+/**
+ * 使用Semaphore的方式，它本质上也是用了“许可”的概念。跟使用LockSupport、BlockingQueue完成此功能是相似的。
+ * 通常来说Semaphore的用法是：多个线程执行一个被同一个semaphore的 semaphore.acquire()和 semaphore.release() 包围的一段代码以控制同时执行这段代码的线程数量【哪个线程acquire了semaphore的许可，哪个线程就release许可】 
+ * 但现在是一个线程中多次执行semaphore.acquire() 导致该semaphore许可用尽而阻塞，然后等待其他的线程中进行该semaphore.release()许可回收。
+ */
+public class Print1A2B3C_Semaphore {
+
+	public static void main(String[] args) {
+		//创建两个Semaphore用于线程间通信
+		Semaphore s1 = new Semaphore(1);
+		Semaphore s2 = new Semaphore(1);
+
+		//t1线程输出的数字
+		char[] aI = "1234567".toCharArray();
+		
+		//t2线程输出的字母
+		char[] aC = "ABCDEFG".toCharArray();
+		
+		
+		Thread t1 = new Thread(() -> {
+			for(char c : aI) {
+				//t1 直接进入循环，先s1.acquire()，第一次循环时不会阻塞
+				try { s1.acquire(); } catch (Exception e) {}
+				
+				System.out.print(c);
+				//t1 输出一个元素后，释放s2，让t2执行
+				try { s2.release(); } catch (Exception e) {}
+			}
+		});
+		
+		Thread t2 = new Thread(() -> {
+			//因为控制t1先输出，所以t2要控制住一开始不能输出，循环执行先s2.acquire()，注意此时不会阻塞
+			try { s2.acquire(); } catch (Exception e) {}
+			
+			//然后进入循环
+			for(char c : aC) {
+				try { s2.acquire(); } catch (Exception e) {}
+				System.out.print(c + " ");
+				try { s1.release(); } catch (Exception e) {}
+			}
+		});
+		
+		//启动两个线程
+		t1.start();
+		t2.start();
+	}
+	
+}