all-of-frontend/Answer/1 ~ 10/3.md
2021-03-11 07:41:19 +08:00

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```js
// 请使用原生代码实现一个Events模块可以实现自定义事件的订阅、触发、移除功能
const fn1 = (... args)=>console.log('I want sleep1', ... args)
const fn2 = (... args)=>console.log('I want sleep2', ... args)
const event = new Events();
event.on('sleep', fn1, 1, 2, 3);
event.on('sleep', fn2, 1, 2, 3);
event.fire('sleep', 4, 5, 6);
// I want sleep1 1 2 3 4 5 6
// I want sleep2 1 2 3 4 5 6
event.off('sleep', fn1);
event.once('sleep', () => console.log('I want sleep'));
event.fire('sleep');
// I want sleep2 1 2 3
// I want sleep
event.fire('sleep');
// I want sleep2 1 2 3
```
今天的题目算是一道常考题了,没有一个标准解法,输出正确就行。
但是如果你能用上一些 ES6 的语法以及处理好一些边界问题,面试官对你的评价会更好点。
```js
class Events {
constructor() {
this.events = new Map();
}
addEvent(key, fn, isOnce, ...args) {
const value = this.events.get(key) ? this.events.get(key) : this.events.set(key, new Map()).get(key)
value.set(fn, (...args1) => {
fn(...args, ...args1)
isOnce && this.off(key, fn)
})
}
on(key, fn, ...args) {
if (!fn) {
console.error(`没有传入回调函数`);
return
}
this.addEvent(key, fn, false, ...args)
}
fire(key, ...args) {
if (!this.events.get(key)) {
console.warn(`没有 ${key} 事件`);
return;
}
for (let [, cb] of this.events.get(key).entries()) {
cb(...args);
}
}
off(key, fn) {
if (this.events.get(key)) {
this.events.get(key).delete(fn);
}
}
once(key, fn, ...args) {
this.addEvent(key, fn, true, ...args)
}
}
```