async/await implicitly returns promise?
I took a look at the spec and found the following information. The short version is that an async function
desugars to a generator which yields Promise
s. So, yes, async functions return promises.
According to the tc39 spec, the following is true:
async function <name>?<argumentlist><body>
Desugars to:
function <name>?<argumentlist>{ return spawn(function*() <body>, this); }
Where spawn
"is a call to the following algorithm":
function spawn(genF, self) {
return new Promise(function(resolve, reject) {
var gen = genF.call(self);
function step(nextF) {
var next;
try {
next = nextF();
} catch(e) {
// finished with failure, reject the promise
reject(e);
return;
}
if(next.done) {
// finished with success, resolve the promise
resolve(next.value);
return;
}
// not finished, chain off the yielded promise and `step` again
Promise.resolve(next.value).then(function(v) {
step(function() { return gen.next(v); });
}, function(e) {
step(function() { return gen.throw(e); });
});
}
step(function() { return gen.next(undefined); });
});
}
Your question is: If I create an async
function should it return a promise or not? Answer: just do whatever you want and Javascript will fix it for you.
Suppose doSomethingAsync
is a function that returns a promise. Then
async function getVal(){
return await doSomethingAsync();
}
is exactly the same as
async function getVal(){
return doSomethingAsync();
}
You probably are thinking "WTF, how can these be the same?" and you are right. The async
will magically wrap a value with a Promise if necessary.
Even stranger, the doSomethingAsync
can be written to sometimes return a promise and sometimes NOT return a promise. Still both functions are exactly the same, because the await
is also magic. It will unwrap a Promise if necessary but it will have no effect on things that are not Promises.
The return value will always be a promise. If you don't explicitly return a promise, the value you return will automatically be wrapped in a promise.
async function increment(num) {
return num + 1;
}
// Even though you returned a number, the value is
// automatically wrapped in a promise, so we call
// `then` on it to access the returned value.
//
// Logs: 4
increment(3).then(num => console.log(num));
Same thing even if there's no return! (Promise { undefined }
is returned)
async function increment(num) {}
Same thing even if there's an await
.
function defer(callback) {
return new Promise(function(resolve) {
setTimeout(function() {
resolve(callback());
}, 1000);
});
}
async function incrementTwice(num) {
const numPlus1 = await defer(() => num + 1);
return numPlus1 + 1;
}
// Logs: 5
incrementTwice(3).then(num => console.log(num));
Promises auto-unwrap, so if you do return a promise for a value from within an async
function, you will receive a promise for the value (not a promise for a promise for the value).
function defer(callback) {
return new Promise(function(resolve) {
setTimeout(function() {
resolve(callback());
}, 1000);
});
}
async function increment(num) {
// It doesn't matter whether you put an `await` here.
return defer(() => num + 1);
}
// Logs: 4
increment(3).then(num => console.log(num));
In my synopsis the behavior is indeed inconsistent with traditional return statements. It appears that when you explicitly return a non-promise value from an async function, it will force wrap it in a promise. I don't have a big problem with it, but it does defy normal JS.
ES6 has functions which don't return exactly the same value as the return
. These functions are called generators.
function* foo() {
return 'test';
}
// Logs an object.
console.log(foo());
// Logs 'test'.
console.log(foo().next().value);