sam*_*ces 6 javascript testing mutex typescript
我已经为 javascript/typescript 编写了一个互斥锁实现,但我正在努力解决如何测试它的问题。这是实现:
class Mutex {
private current = Promise.resolve();
async acquire() {
let release: () => void;
const next = new Promise<void>(resolve => {
release = () => { resolve(); };
});
const waiter = this.current.then(() => release);
this.current = next;
return await waiter;
}
}
Run Code Online (Sandbox Code Playgroud)
用法:
const mut = new Mutex();
async function usesMutex() {
const unlock = await mut.acquire();
try {
await doSomeStuff();
await doOtherStuff();
} finally {
unlock();
}
}
Run Code Online (Sandbox Code Playgroud)
如果互斥锁没有按预期工作,我不确定是否有任何简单的方法可以创建会导致测试失败的时序问题。任何建议将不胜感激。
您需要一个确定性的测试,如果互斥锁中断,测试的行为将会改变。
下面的例子是一个原子计数器问题。产生了两个工人,每个工人在循环中做三件事:
至关重要的是,我正在使用await
and setTimeout
here 在执行中创建中断。async
没有任何await
s的函数将是完全原子的,因此我们需要创建一些中断以允许调度程序在任务之间切换。如果互斥锁被破坏,这些await
s 将允许调度程序运行其他工作程序的代码,因为每个await
都是 Javascript 调度程序更改作业的机会。
如果互斥锁正常工作,您应该会看到以下内容。在每一步之间,worker 进入睡眠状态并醒来,但互斥锁不允许其他 worker 做任何事情:
- 工人 1 获取锁
- Worker 1
0
从全局计数器中读取值- 工人 1 将值从
0
到1
- 工人 1 将值写
1
回全局计数器- 工人 1 释放锁
- Worker 2 获取锁
- Worker 2
1
从全局计数器读取值- 工人 2 将值从
1
到2
- Worker 2 将值写
2
回全局计数器- Worker 2 释放锁
结果:2
,预期:2
如果互斥锁不起作用(或未被使用),两个工人就会互相绊倒,最终结果将是不正确的。和以前一样,worker 每次执行一个动作时都会进入睡眠状态:
- Worker 1
0
从全局计数器中读取值- Worker 2
0
从全局计数器读取值- 工人 1 将值从
0
到1
- 工人 2 将值从
0
到1
- 工人 1 将值写
1
回全局计数器- Worker 2 将值写
1
回全局计数器
结果:1
,预期:2
在这两种情况下,两个工人都在执行相同的操作,但如果不执行该命令,则结果是不正确的。
这个例子是人为的,但可重现且主要是确定性的。当互斥体工作时,您将始终获得相同的最终结果。如果不是,您将始终得到错误的结果。
工作演示:
var state = {
isMutexBroken: false,
counter: 0,
worker1LastAction: '',
worker2LastAction: '',
worker1IsActive: false,
worker2IsActive: false,
}
class Mutex {
constructor() {
this.current = Promise.resolve();
}
async acquire() {
if (state.isMutexBroken) {
return () => {};
}
let release;
const next = new Promise(resolve => {
release = () => {
resolve();
};
});
const waiter = this.current.then(() => release);
this.current = next;
return await waiter;
}
}
var mutex = new Mutex();
const renderState = () => {
document.getElementById('mutex-status').textContent = state.isMutexBroken ? 'Mutex is *not* working correctly. Press "fix mutex" to fix it.' : 'Mutex is working correctly. Press "break mutex" to break it.';
document.getElementById('counter').textContent = `Counter value: ${state.counter}`;
document.getElementById('worker1').textContent = `Worker 1 - last action: ${state.worker1LastAction}`;
document.getElementById('worker2').textContent = `Worker 2 - last action: ${state.worker2LastAction}`;
document.getElementById('start-test').disabled = state.worker1IsActive || state.worker2IsActive;
document.getElementById('break-mutex').disabled = state.worker1IsActive || state.worker2IsActive;
document.getElementById('fix-mutex').disabled = state.worker1IsActive || state.worker2IsActive;
}
// https://stackoverflow.com/a/39914235/8166701
function sleep(ms) {
return new Promise(resolve => setTimeout(resolve, ms));
}
const worker = async(delay, count, id) => {
state[`${id}IsActive`] = true;
let workerCopyOfCounter;
for (let i = 0; i < count; i++) {
const unlock = await mutex.acquire();
state[`${id}LastAction`] = `Aquired lock.`;
renderState();
await sleep(delay);
workerCopyOfCounter = state.counter;
state[`${id}LastAction`] = `Acquired global counter: ${workerCopyOfCounter}`;
renderState();
await sleep(delay);
workerCopyOfCounter++;
state[`${id}LastAction`] = `Incremented counter: ${workerCopyOfCounter}`;
renderState();
await sleep(delay);
state.counter = workerCopyOfCounter;
state[`${id}LastAction`] = `Wrote ${workerCopyOfCounter} back to global counter.`;
renderState();
await sleep(delay);
unlock();
state[`${id}LastAction`] = `Released lock.`;
renderState();
await sleep(delay);
}
state[`${id}LastAction`] = `Finished.`;
state[`${id}IsActive`] = false;
renderState();
}
document.getElementById('break-mutex').onclick = () => {
state.isMutexBroken = true;
renderState();
}
document.getElementById('fix-mutex').onclick = () => {
state.isMutexBroken = false;
renderState();
}
document.getElementById('start-test').onclick = () => {
document.getElementById('test-result').textContent = '';
document.getElementById('start-test').textContent = 'Reset and start test';
state.counter = 0;
state.worker1LastAction = '';
state.worker2LastAction = '';
renderState();
const slow = document.getElementById('slow').checked;
const multiplier = slow ? 10 : 1;
Promise.all([
worker(20 * multiplier, 10, 'worker1'),
worker(55 * multiplier, 5, 'worker2')
]).then(() => {
const elem = document.getElementById('test-result');
elem.classList.remove('pass');
elem.classList.remove('fail');
elem.classList.add(state.counter === 15 ? 'pass' : 'fail');
elem.textContent = state.counter === 15 ? 'Test passed' : 'Test failed';
});
}
renderState();
Run Code Online (Sandbox Code Playgroud)
.flex-column {
display: flex;
flex-direction: column;
}
.flex-row {
display: flex;
}
.top-padding {
padding-top: 8px;
}
.worker-state-container {
background-color: #0001;
margin-top: 8px;
padding: 5px;
}
.pass {
background-color: limegreen;
color: white;
}
.fail {
background-color: red;
color: white;
}
Run Code Online (Sandbox Code Playgroud)
<div class="flex-column">
<div className="flex-row">
<button id="break-mutex">Break mutex</button>
<button id="fix-mutex">Fix mutex</button>
<div id="mutex-status"></div>
</div>
<div className="flex-row">
<input type="checkbox" id="slow" name="slow"><label for="slow">slow</label>
</div>
<div class="flex-row top-padding">
<button id="start-test">Start test</button>
</div>
<div id="counter"></div>
<div>Expected end value: 15</div>
<div id="test-result"></div>
<div class="top-padding">
<div id="worker1" class="worker-state-container">
</div>
<div id="worker2" class="worker-state-container">
</div>
</div>
</div>
Run Code Online (Sandbox Code Playgroud)
最小版本:
var state = { counter: 0 }
// https://stackoverflow.com/a/39914235/8166701
function sleep(ms) {
return new Promise(resolve => setTimeout(resolve, ms));
}
const worker = async (delay, count) => {
let workerCopyOfCounter;
for (let i = 0; i < count; i++) {
// Lock the mutex
const unlock = await mutex.acquire();
// Acquire the counter
workerCopyOfCounter = state.counter;
await sleep(delay);
// Increment the local copy
workerCopyOfCounter++;
await sleep(delay);
// Write the local copy back to the global counter
state.counter = workerCopyOfCounter;
await sleep(delay);
// Unlock the mutex
unlock();
await sleep(delay);
}
}
// Create two workers with different delays. If the code is working,
// state.counter will equal 15 when both workers are finished.
Promise.all([
worker(20, 10),
worker(55, 5),
]).then(() => {
console.log('Expected: 15');
console.log('Actual:', state.counter);
});
Run Code Online (Sandbox Code Playgroud)
归档时间: |
|
查看次数: |
403 次 |
最近记录: |