If I have an async function, do I have to repeatedly use the await keyword down the call chain, even if the methods being invoked are marked as async and use the await keyword themselves?
async function foo() {
const result = await bar(); // is this await required?
console.log(result);
}
async function bar() {
console.log('waiting...');
return await new Promise(resolve => setTimeout(() => resolve('result'), 1000)) ;
}
You do if you want the promise to be resolved.
Because
const result = bar();
is perfectly valid: it returns a promise.
Using await, you instruct the engine to wait for the promise to be resolved (or fail) and get the result. There are many valid cases where you would want to deal with the promise itself (for example adding operations) even inside an async function.
No you dont:
async function foo() {
const result = await baz(); //this is just required
console.log(result);
}
async function baz(){//mustnt be async
return bar();//simply return, no await required
}
async function bar() {//mustnt be async
console.log('waiting...');
return new Promise(resolve => setTimeout(() => resolve('result'), 1000)) ;//this isnt
}
You just need to await the promise once at the highest level ( if youre not planning to change the data somewhere).
The upper example does not need to be async , however this has to:
async function buy(product){
if(await isAvailableNow()){
return buy();//buy is a promise
}else{
return buySomewhereElse();//another promise
}
}
If I have an async function, do I have to repeatedly use the await keyword down the call chain
If the function is async and you want to run your instructions (that follows the call) AFTER the async function is done, then yes. Always. You have used
return await new Promise(...);
in bar() which is a strange thing. The await is not required here because you have defined the function as async function bar() { ... }. It implicitly returns with a Promise.resolve(...) object. So at the end, no matter of what you do in the return statement, you always get that object.
So when calling bar() as
const result = bar();
// stmt
then result contains a promise object which is doing the task in bar() function asynchronously AND runs the stmt after it. So the stmt is executed, even if the bar() function is not done yet.
To resolve it, you have to use await at the call...
const result = await bar();
// stmt
In this situation, stmt is only executed if bar is done.
If you want the result of the Promise before solving the async you'll need the await. The only function of the await is wait a Promise to be solved, but await is not mandatory in async functions, is just for that case. Just think "I have a Promise? I need this solved before the async?" If the answer was a double yes, so use await.
Related
When I write an async function it usually returns a promise:
export const myPromiseFunction = async (params) => {
// some logic
return Promise.resolve('resolved-value');
});
But I was wondering if it would be a mistake if this function would not return a promise, so for example:
export const myPromiseFunction = async (params) => {
// some logic
params.map(async (param) => {
await printParam(param);
async function printParam(par) {
// do some other stuff
Printer.print(par);
});
});
});
export class Printer {
public static async print(par) {console.log(par);} // I know it could not be async, but for the sake lets suppose it does
}
Is this a mistake / bad practice ? Or can we find a scenario when this will be valid and desirable ?
All async functions automatically return Promises. If you declare a function as async, it will return a Promise, even if your only return value is a simple value like a string or number. If you don't return explicitly, your async function will still return a Promise with a value of undefined.
In fact, it is more common for the body of an async function to return a simple value rather than a promise; the assumption is that your function is async because you await the promises you consume within it. As a consequence, even if you return the value 5, the return value is a Promise (that resolves to 5) representing the potential delay that comes from any await expressions in the function.
You don't have to return a Promise object explicitly in your async function, and it is redundant to do so if you're just wrapping a simple value like 'resolved-value'. Conversely, you can make a normal function behave like an async function if you always return a Promise (potentially with Promise.resolve) and you never synchronously throw an error within it.
async function myFunction() {
makeSomeCall(); // Happens immediately.
await someOtherPromise(); // myFunction returns a Promise
// before someOtherPromise resolves; if it
// does without error, the returned Promise
return 5; // resolves to 5.
}
/** Behaves the same as the above function. */
function myFunctionButNotAsync() {
try {
makeSomeCall();
// If you didn't have someOtherPromise() to wait for here, then
// this is where Promise.resolve(5) would be useful to return.
return someOtherPromise().then(() => 5);
} catch (e) {
return Promise.reject(e);
}
}
All that said, you may have occasion to explicitly return a Promise object (such as one produced by Promise.all or a separate Promise-returning function), which then observes rules similar to what Promise.resolve() observes: If the object you return from an async function is a Promise, or has a then function, then the automatic Promise the async function returns will wait for the specific Promise or Promise-like object you pass back with return.
async function myFunction() {
makeSomeCall(); // Happens immediately.
await anythingElse(); // You can still await other things.
return someOtherPromise(); // The promise myFunction returns will take
// the same outcome as the Promise that
// someOtherPromise() returns.
}
In a related sense, this is why return await is seen as redundant, though as described it does make a difference for the stack traces that you see if the wrapped promise is rejected.
Short answer: no, an async function doesn't have to returns a Promise. Actually, generally you wouldn't return a Promise object (unless you're chaining asynchronous events).
What async and await do is wait for a response from something that returns a Promise.
You first code example actually returns a resolved Promise. But what happens if the Promise isn't resolved properly ?
It's best to call a function that returns a Promise from another async function:
function getRequestResult() {
return new Promise(resolve => {
setTimeout(() => {
resolve('request sent');
}, 2000);
});
}
async function sendMyRequest() {
console.log('Sending request');
const result = await getRequestResult();
console.log(result);
// expected output: "resolved"
}
You can send rejections/errors within getRequestResult() that way, and also manage how these errors will be managed by the call in sendMyRequest() (see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/await).
I have two functions.
1.
async function firstFunction() {
// Do stuff here
// I can't just return doSomeOtherThings because the firstFunction
// returns different data then what doSomeOtherThings does
doSomeOtherThings();
return;
}
async function doSomeOtherThings() {
// do promise stuff here
// this function runs some db operations
}
If I run firstFunction(); will it execute my doSomeOtherThings() function, or will it return early from that and cause some or all of the doSomeOtherThings code to not be executed? Do I need to do await doSomeOtherThings()?
I think there is a bit of confusion here, so I'll try to start from the beggining.
First, async functions always return a promise. If you add another async function inside that, you COULD chain them and wait for a response before returning the first promise. However, if you dont await the inner function, the first function will be resolved while the second is still running.
async function firstFunction() {
if (I want to wait for doSomeOtherThings to finished before ending firstFunction){
await doSomeOtherThings();
} else if (I can finish firstFUnction and let doSomeOtherTHings finish later){
doSomeOtherThings
}
return;
}
async function init() {
const apiResponse = await firstFunction();
};
init();
You need to have await. Actually 'async/await' is just a syntax sugar.
You function updateOtherThings actually returns a Promise, and if you don't await it, then it simply doesn't run.
If you want simply to start and forget it, then write something like:
updateOtherThings().then(() => {});
This question already has answers here:
Async function without await in JavaScript
(4 answers)
Closed 2 years ago.
due to everyone's help, I got the logic behind async, promise, then, await.
I have one curisosity on top of its basic nature, which is what if
I declare async function but doesn't use await in it.
technically every argument within async function is invisibly capsulated by '.then()'
but how it works would be exactly same as synchronous function execution.
For example,
async function trackUserHandler() {
var positiondata= await getPosition();
var timerdata = await setTimer(2000)
console.log(positiondata, timerdata);
setTimer(1000).then(() => {
console.log('Timer done!');
});
console.log('one');
}
The console below doesn't run till the first two await function is done due to await(s) sitting before this.
console.log(positiondata, timerdata);
What if I don't put any await(s) in the async like below?
async function trackUserHandler() {
var positiondata= getPosition();
var timerdata = setTimer(2000)
console.log(positiondata, timerdata);
setTimer(1000).then(() => {
console.log('Timer done!');
});
console.log('one');
}
I test-run this code and behaves seemigly same as regular function without 'async'.
Of course behind the scene, everything in the function is encapsulated into 'then()' though.
Am I understanding right?
Thank you in advance.
Behind the scene- If I use Aync but doesn't use await in it, would it be identical with normal function?
Yes, you are right. An async function without an await expression will run synchronously and would be identical with normal function, but the only difference is that async functions always return a promise. If the return value of an async function is not explicitly a promise, it will be implicitly wrapped in a promise.
For example, the following:
async function foo() {
return 1
}
...is equivalent to:
function foo() {
return Promise.resolve(1)
}
If there is an await expression inside the function body, however, the async function will always complete asynchronously.
For example:
async function foo() {
await 1
}
...is equivalent to:
function foo() {
return Promise.resolve(1).then(() => undefined)
}
Code after each await expression can be thought of as existing in a .then callback.
Yes, an async function runs synchronously till the first await, so it behaves like a regular function. It does return a Promise though:
function trackUserHandler() {
// ... code
return Promise.resolve(undefined);
}
In your example, the 2 functions won't behave the same. Without the await keyword, your variables won't capture the results returned by these 2 functions, but instead receive 2 Promises.
var positiondata = getPosition();
var timerdata = setTimer(2000);
So your console.log will print out 2 Promises instead of the values you actually expect.
An async function can contain an await expression, that pauses the
execution of the async function and waits for the passed Promise's
resolution, and then resumes the async function's execution and
returns the resolved value.
As you assumed, if no await is present the execution is not paused and your code will then be executed in a non-blocking manner.
const getPromise = async (s) => {
return new Promise((resolve, reject) => {
setTimeout(() => resolve(s), 500);
});
}
(async() => {
try {
const result = getPromise("a"); //no await, result has not been unwrapped
console.log('async/await -> ', result);
} catch (err) {
console.log(err);
}
})();
Async function without await in Javascript
I have an async method for example async methodA(),
I want methodB() to not be async but to call methodA().
How do I get methodA() to finish and send data to methodB()?
You can use async/await .
Let's take your method A which is async. Make your method A async by using async keyword.
async function A(){
//do something
}
Now function A always return a promise if it doesn't return JavaScript wraps it around into a resolved promise
Make your method B to wait until that promise settles and return it's result
let value = await promise;
Now this only works inside async function. And you can use value in method B.
With this you can use the output of method A in method BYou can also assign your output to a global variable and use it later
async function A() {
let promise = new Promise((resolve, reject) => {
setTimeout(() => resolve("output for method B"), 1000)
});
let result = await promise; // wait till the promise resolves (*)
function B(){
// do something with result
console.log(result)
}
B(result)
}
A();
Well, It's not exactly possible without a timeout and I didn't want one of those so I made the whole thing async...
I'm trying to learn async-await. In this code -
const myFun = () => {
let state = false;
setTimeout(() => {state = true}, 2000);
return new Promise((resolve, reject) => {
setTimeout(() => {
if(state) {
resolve('State is true');
} else {
reject('State is false');
}
}, 3000);
});
}
const getResult = async () => {
return await myFun();
}
console.log(getResult());
why am I getting output as -
Promise { <pending> }
Instead of some value? Shouldn't the getResult() function wait for myFun() function resolve it's promise value?
If you're using async/await, all your calls have to use Promises or async/await. You can't just magically get an async result from a sync call.
Your final call needs to be:
getResult().then(response => console.log(response));
Or something like:
(async () => console.log(await getResult()))()
What you need to understand is that async/await does not make your code run synchronously, but let's you write it as if it is:
In short: The function with async in front of it is literally executed asynchronously, hence the keyword "async". And the "await" keyword wil make that line that uses it inside this async function wait for a promise during its execution. So although the line waits, the whole function is still run asynchronously, unless the caller of that function also 'awaits'...
More elaborately explained: When you put async in front of a function, what is actually does is make it return a promise with whatever that function returns inside it. The function runs asynchronously and when the return statement is executed the promise resolves the returning value.
Meaning, in your code:
const getResult = async () => {
return await myFun();
}
The function "getResult()" will return a Promise which will resolve once it has finished executing. So the lines inside the getResult() function are run asynchronously, unless you tell the function calling getResult() to 'await' for it as well. Inside the getResult() function you may say it must await the result, which makes the execution of getResult() wait for it to resolve the promise, but the caller of getResult() will not wait unless you also tell the caller to 'await'.
So a solution would be calling either:
getResult().then(result=>{console.log(result)})
Or when using in another function you can simply use 'await' again
async function callingFunction(){
console.log(await(getResult());
}
This is my routine dealing with await and async using a Promise with resolve and reject mechanism
// step 1 create a promise inside a function
function longwork()
{
p = new Promise(function (resolve, reject) {
result = 1111111111111 // long work here ;
if(result == "good"){
resolve(result);
}
else
{
reject("error ...etc")
}
})
return p
}
// step 2 call that function inside an async function (I call it main)and use await before it
async function main()
{
final_result = await longwork();
//..
}
//step 3 call the async function that calls the long work function
main().catch((error)=>{console.log(error);})
Hope that saves someone valuable hours
What hasn't been mentioned in this discussion are the use-case implications of the behaviour. The key thing, as I see it, is to consider what you are planning to do with the output from the top level, truly asynchronous function, and where you are planning to do that.
If you are planning to consume the output immediately, i.e. within the "async" function that is awaiting the return of the top level asynchronous function, and what you do with the output has no implication for other functions deeper in the call stack, then it does not matter that the deeper functions have moved on. But if the output is needed deeper in the call stack, then you need use "async" functions making await calls all the way down the stack to that point. Once you reach a point in the call stack where the function does not care about the asynchronous output, then you can stop using async functions.
For example, in the following code, function B uses the stuff returned from function A so is declared "async" and awaits A(). Function C() calls B(), is returned a Promise, but can move straight on before that promise is resolved because it is not interested in A()'s stuff, nor what's done with it. So C does not need to be declared as async, nor await B().
function A() {
return new Promise((resolve, reject) => {
//do something slow
resolve (astuff)
}
}
async function B() {
var bstuff = await A();
dosomethingwith(bstuff);
return;
}
function C() {
B();
dontwaitmoveon();
...
return;
}
In this next example, C() does use A()'s stuff, so needs to wait for it. C() must be declared "async" and await B(). However D() does not care about A()'s stuff, nor what's done with it, so moves on once C() returns its promise.
function A() {
return new Promise((resolve, reject) => {
//do something slow
resolve (astuff)
}
}
async function B() {
var bstuff = await A();
dosomething();
return bstuff;
}
async function C() {
var cstuff = await B();
dosomethingwith(cstuff);
...
return;
}
function D() {
C();
dontwaitmoveon();
...
return;
}
Since figuring this out, I have tried to design my code so the stuff returned by the asynchronous function is consumed as close as possible to the source.
Though your "getResult" function is async and you have rightly made an await call of myFun, look at the place where you call the getResult function, it is outside any async functions so it runs synchronously.
So since getResult called from a synchronous point of view, as soon as it is called, Javascript synchronously gets whatever result is available at the moment, which is a promise.
So returns from an async function cannot be forced to await(very important), since they are synchronously tied to the place of origin of the call.
To get what you want you can run the below,
async function getResult() {
const result = await myFun();
console.log(result);
//see no returns here
}
getResult();