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const q = () => console.log(2) const w = async () => { console.log(1) await q console.log(3) } w()
为什么q没有输出 1 2 3而是输出 1 3
小伙看你根骨奇佳，潜力无限，来学PHP伐。
低级错误，函数执行应该加上括号 await q();。
await q();
![[Web front-end] Node.js quick start](https://img.php.cn/upload/course/000/000/067/662b5d34ba7c0227.png)
低级错误，函数执行应该加上括号
await q();。