gpt4 book ai didi

java - 为多线程 FizzBu​​zz 编写测试用例

转载 作者:行者123 更新时间:2023-12-01 12:47:47 25 4
gpt4 key购买 nike

我正在通过此链接解决 LeetCode 问题:https://leetcode.com/problems/fizz-buzz-multithreaded/ .
基本上,我正在编写一个执行标准 fizzbuzz 的 FizzBu​​zz 类,但同时运行 4 个线程,每个线程调用不同的方法。 (一个线程正在调用 fizzbuzz,一个线程在调用 fizz,一个调用 buzz 和一个调用号码)。这是我接受的解决方案:

class FizzBuzz {
private int n;
int num = 1; // FizzBuzz starts at 1.
public FizzBuzz(int n) {
this.n = n;
}

// printFizz.run() outputs "fizz".
public synchronized void fizz(Runnable printFizz) throws InterruptedException {
while(num <= n) {
if(num % 3 == 0 && num % 5 != 0) {
printFizz.run();
num++;
notifyAll();
} else {
wait();
}
}
}

// printBuzz.run() outputs "buzz".
public synchronized void buzz(Runnable printBuzz) throws InterruptedException {
while(num <= n) {
if(num % 3 != 0 && num % 5 == 0) {
printBuzz.run();
num++;
notifyAll();
} else {
wait();
}
}
}

// printFizzBuzz.run() outputs "fizzbuzz".
public synchronized void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {
while(num <= n) {
if(num % 15 == 0) {
printFizzBuzz.run();
num++;
notifyAll();
} else {
wait();
}
}
}



// printNumber.accept(x) outputs "x", where x is an integer.
public synchronized void number(IntConsumer printNumber) throws InterruptedException {
while(num <= n) {
if(num % 3 != 0 && num % 5 != 0) {
printNumber.accept(num);
num++;
notifyAll();
} else {
wait();
}
}
}
}
问题是我很难在我的 IDE 中模拟这种情况。
这是我的主要方法:
    public class FizzBuzzMain {
public static void main(String[] args) {
FizzBuzz fizzBuzz = new FizzBuzz(15);
Runnable printFizzBuzz = new PrintFizzBuzz();
Runnable printFizz = new PrintFizz();
Runnable printBuzz = new PrintBuzz();

Thread t1 = new Thread(printFizzBuzz);
Thread t2 = new Thread(printFizz);
Thread t3 = new Thread(printBuzz);
IntConsumer printNumber = new PrintNumber();

t1.start();
t2.start();
t3.start();

try {
fizzBuzz.number(printNumber);
fizzBuzz.fizz(printFizz);
fizzBuzz.buzz(printBuzz);
fizzBuzz.buzz(printFizzBuzz);
} catch(Exception e) {
e.printStackTrace();
}
}
}

我想用 IntConsumer 初始化第四个线程(这是从问题中给出的),这样我也可以执行 t4 ,但我不知道该怎么做。显然,我的 main 方法在某个时刻停止运行,因为所有线程都进入等待状态并且没有人唤醒它们(此处缺少 t4)。
任何帮助将不胜感激!

最佳答案

我修复了您的类,以便您可以在 IDE 中运行它,然后实现了 Emma 提供的解决方案:

abstract class FizzBuzzRunner implements Runnable {

protected FizzBuzz fizzBuzz;

public FizzBuzzRunner(FizzBuzz fizzBuzz) {
this.fizzBuzz = fizzBuzz;
}

abstract protected void print();
}

class PrintFizz extends FizzBuzzRunner {

public PrintFizz(FizzBuzz fizzBuzz) {
super(fizzBuzz);
}

@Override
public void run() {
try {
this.fizzBuzz.fizz(this);
} catch (InterruptedException e) {
e.printStackTrace();
}
}

@Override
protected void print() {
System.out.println("fizz");
}
}

class PrintBuzz extends FizzBuzzRunner {

public PrintBuzz(FizzBuzz fizzBuzz) {
super(fizzBuzz);
}

@Override
public void run() {
try {
this.fizzBuzz.buzz(this);
} catch (InterruptedException e) {
e.printStackTrace();
}
}

@Override
protected void print() {
System.out.println("buzz");
}
}

class PrintFizzBuzz extends FizzBuzzRunner {

public PrintFizzBuzz(FizzBuzz fizzBuzz) {
super(fizzBuzz);
}

@Override
public void run() {
try {
this.fizzBuzz.fizzbuzz(this);
} catch (InterruptedException e) {
e.printStackTrace();
}
}

@Override
protected void print() {
System.out.println("fizzbuzz");
}
}

class IntConsumer extends PrintFizzBuzz {

public IntConsumer(FizzBuzz fizzBuzz) {
super(fizzBuzz);
}

@Override
public void run() {
try {
this.fizzBuzz.number(this);
} catch (InterruptedException e) {
e.printStackTrace();
}
}

public void accept(int n) {
System.out.println(n);
}
}
对主要的一个小改动:
public class Main {
public static void main(String[] args) {
FizzBuzz fizzBuzz = new FizzBuzz(15);
Runnable printFizzBuzz = new PrintFizzBuzz(fizzBuzz);
Runnable printFizz = new PrintFizz(fizzBuzz);
Runnable printBuzz = new PrintBuzz(fizzBuzz);
Runnable printNumber = new IntConsumer(fizzBuzz);

Thread t1 = new Thread(printFizzBuzz);
Thread t2 = new Thread(printFizz);
Thread t3 = new Thread(printBuzz);
Thread t4 = new Thread(printNumber);

t1.start();
t2.start();
t3.start();
t4.start();
}
}
最后是 FizzBu​​zz 类:
class FizzBuzz {
private int n;
private Semaphore semNumber;
private Semaphore semFizz;
private Semaphore semBuzz;
private Semaphore semFizzBuzz;

public FizzBuzz(int n) {
this.n = n;
semNumber = new Semaphore(1);
semFizz = new Semaphore(0);
semBuzz = new Semaphore(0);
semFizzBuzz = new Semaphore(0);
}

// printFizz.print() outputs "fizz".
public void fizz(FizzBuzzRunner printFizz) throws InterruptedException {
for (int counter = 3; counter <= n; counter += 3) {
semFizz.acquire();
printFizz.print();

if ((counter + 3) % 5 == 0) {
counter += 3;
}

semNumber.release();
}
}

// printBuzz.print() outputs "buzz".
public void buzz(FizzBuzzRunner printBuzz) throws InterruptedException {
for (int counter = 5; counter <= n; counter += 5) {
semBuzz.acquire();
printBuzz.print();

if ((counter + 5) % 3 == 0) {
counter += 5;
}

semNumber.release();
}
}

// printFizzBuzz.print() outputs "fizzbuzz".
public void fizzbuzz(FizzBuzzRunner printFizzBuzz) throws InterruptedException {
for (int counter = 15; counter <= n; counter += 15) {
semFizzBuzz.acquire();
printFizzBuzz.print();
semNumber.release();
}
}

// printNumber.accept(x) outputs "x", where x is an integer.
public void number(IntConsumer printNumber) throws InterruptedException {
for (int counter = 1; counter <= n; counter++) {
semNumber.acquire();

if (counter % 15 == 0) {
semFizzBuzz.release();

} else if (counter % 5 == 0) {
semBuzz.release();

} else if (counter % 3 == 0) {
semFizz.release();

} else {
printNumber.accept(counter);
semNumber.release();
}
}
}
}

关于java - 为多线程 FizzBu​​zz 编写测试用例,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62786471/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com