67 lines
1.7 KiB
Markdown
67 lines
1.7 KiB
Markdown
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```js
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// 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
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const fn1 = (... args)=>console.log('I want sleep1', ... args)
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const fn2 = (... args)=>console.log('I want sleep2', ... args)
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const event = new Events();
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event.on('sleep', fn1, 1, 2, 3);
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event.on('sleep', fn2, 1, 2, 3);
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event.fire('sleep', 4, 5, 6);
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// I want sleep1 1 2 3 4 5 6
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// I want sleep2 1 2 3 4 5 6
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event.off('sleep', fn1);
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event.once('sleep', () => console.log('I want sleep'));
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event.fire('sleep');
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// I want sleep2 1 2 3
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// I want sleep
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event.fire('sleep');
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// I want sleep2 1 2 3
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```
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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。
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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。
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```js
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class Events {
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constructor() {
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this.events = new Map();
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}
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addEvent(key, fn, isOnce, ...args) {
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const value = this.events.get(key) ? this.events.get(key) : this.events.set(key, new Map()).get(key)
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value.set(fn, (...args1) => {
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fn(...args, ...args1)
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isOnce && this.off(key, fn)
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})
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}
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on(key, fn, ...args) {
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if (!fn) {
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console.error(`没有传入回调函数`);
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return
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}
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this.addEvent(key, fn, false, ...args)
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}
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fire(key, ...args) {
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if (!this.events.get(key)) {
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console.warn(`没有 ${key} 事件`);
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return;
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}
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for (let [, cb] of this.events.get(key).entries()) {
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cb(...args);
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}
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}
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off(key, fn) {
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if (this.events.get(key)) {
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this.events.get(key).delete(fn);
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}
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}
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once(key, fn, ...args) {
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this.addEvent(key, fn, true, ...args)
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}
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}
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```
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