Careercup - Microsoft面试题 - 6366101810184192 - Java

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2014-05-10 22:30

题目链接

原题：

Design database locks to allow r/w concurrency and data consistency.

题目：设计一个支持并行读写并保证数据一致性的数据库锁。

解法：这是什么范畴的面试题？对于我这种完全没有相关项目经验的人，还真是无从下手。翻了书之后顺便重新复习了一下共享锁、互斥锁的概念。说白了，这些都是课本上的基础知识，可是毕业没多久就忘光了。看来知识不用，保质期比水果还短。然后我琢磨着锁的模型，写了个模拟的代码。这个代码和SQL毫无关系，可能也和正确答案相去甚远。姑妄言之，姑妄听之吧。

代码：

// http://www.careercup.com/question?id=6366101810184192
import java.util.concurrent.Semaphore; public class FooBar {
public static final int MAX_CONCURRENT_READ = 100;
private Semaphore sharedSemaphore;
private Semaphore exclusiveSemaphore; public FooBar() {
// TODO Auto-generated constructor stub
this.sharedSemaphore = new Semaphore(MAX_CONCURRENT_READ);
this.exclusiveSemaphore = new Semaphore(1);
} public void reader() {
// The reader here is not to return a value, but to perform read()
// action. Thus it is 'void reader()'.
while (exclusiveSemaphore.availablePermits() < 1) {
try {
Thread.sleep(50);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
} try {
sharedSemaphore.acquire();
System.out.println("Performing read() operation.");
sharedSemaphore.release();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
} public void writer() {
while (exclusiveSemaphore.availablePermits() < 1
&& sharedSemaphore.availablePermits() < MAX_CONCURRENT_READ) {
try {
Thread.sleep(50);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
} try {
exclusiveSemaphore.acquire();
System.out.println("Performing write() operation.");
exclusiveSemaphore.release();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
} public static void main(String[] args) {
FooBar fooBar = new FooBar(); fooBar.reader();
fooBar.writer();
}
}