为什么异步函数在等待已解决的承诺时返回？

Question

Indeed, such behaviour can cause some performance issuses, ie additional context switches between caller "thread" and "continuation tail" of async function.

 async function f() { await Promise.resolve("something"); console.log("f: after await"); return "somthing_else"; } function g() { const fres = f(); console.log("g: f returns"); fres.then((x) => console.log("finish")); } g();

The output is (at least in my Chrome):

g: f returns
f: after await
finish

Since await operator actually has nothing to wait, I would expect:

f: after await
g: f returns
finish

Is that behaviour specified by standard or it is implementation dependent?

If it is standard behaviour what is the reason?

Answer 1

f function is async, so you should await for it within g function

async function g(){
  const fres = await f();
  console.log("g: f returns");
  fres.then((x) => console.log("finish"));
}

Answer 2

Promises are placed in a queue that is executed after the current normal script queue is empty.

So you will finish running g() and hence, push console.log("g: f returns"); to the current stack before f() resolves.

If the log was inside the fres.then() function, it would only be logged after f() resolves.