异步/等待清晰，以睡眠为例

我试图通过下面的实现来掌握 async/await 的窍门，但它没有按预期工作

    public static async sleep(ms: number): Promise<void> {
        await Utilities._sleep(ms);
    }

    private static _sleep(ms: number): Promise<{}> {
        return new Promise((resolve: Function) => setTimeout(resolve, ms));
    }

_sleep将在 n 毫秒后解决承诺，并且等待应该sleep直到那个时候..

但我的下面的测试失败了

it("should sleep for 500 ms", ()=> {
    const date1 = (new Date()).getTime();
    Utilities.sleep(500);
    const date2 = (new Date()).getTime();
    chai.expect(date2 - date1).to.least(500);
})

有消息

 sleep should sleep for 500 ms FAILED
    AssertionError: expected 2 to be at least 500

我的理解是：sleep将等待直到承诺_sleep已解决（根据 setTimeout 将在 500 毫秒后解决）

EDIT

测试框架在mocha

你还没有等到你的sleep()调用（如用户@igor在您问题的评论中所述）：

简化版：

async function sleep(ms: number) {
    await _sleep(ms);
}

function _sleep(ms: number) {
    return new Promise((resolve) => setTimeout(resolve, ms));
}

console.time('start')
sleep(500).then(() => {
    console.timeEnd('start')
})

// If your test runner supports async:
it("should sleep for 500 ms", async () => {
    const date1 = (new Date()).getTime();
    await sleep(500);
    const date2 = (new Date()).getTime();
    chai.expect(date2 - date1).to.least(500);
})