I have below code:
let func = () => {
return new Promise((resolve) => {
setTimeout(() => {
console.log("two");
resolve();
}, 3000);
})
};
let func2 = async () => {
console.log("one");
await func();
console.log("three");
}
(async () => {
await func2();
console.log("main"); // This should never be executed
})()
Noticed func2 never returns a value, the promise returned by func2 should never be fullfilled in my opinion. Hence console.log("main") should never be executed. However it is executed after console.log("three"). Can somebody explain this to me?
Noticed func2 never returns a value, the promise returned by func2 should never be fullfilled in my opinion.
That is not how async functions work. Your func2 returns when it is done executing. The return value of a function with no return statement is the specific value undefined. So, undefined becomes the resolved value of the promise. Remember, in Javascript, undefined is a specific value. It's as if you did return undefined at the end of your function block. So, since undefined is the return value, that becomes the resolved value of the async promise.
To fully cover all the bases, async functions always return a promise and that promise gets a resolved/rejected value one of these ways:
1. When you explicitly return a value from the function. That becomes the resolved value of the promise that the async function returns.
async function test() {
return "hello";
}
test().then(val => {
console.log(val); // "hello"
});
2. When you throw an exception. The exception becomes the reject reason of the promise that the async function returns.
async function test() {
throw new Error("ouch");
}
test().catch(err => {
console.log(err); // Shows error object with message "ouch"
});
3. When you return another promise. That promise is chained to the one that the async function returns and the promise that the async function returns will follow the one you returned (resolve when it resolves with the same value or reject with the same reason).
async function test() {
return new Promise(resolve => {
setTimeout(() => {
resolve("hi");
}, 1000);
});
}
test().then(val => {
console.log(val); // "hi"
});
4. When you return nothing. That is the same as in regular functions and it's equivalent to a return undefined at the end of the function block so the resolved value takes on the value of undefined.
So, this:
async function test() {
console.log("hi");
}
test().then(val => {
console.log(val); // undefined
});
Works exactly the same as this:
async function test() {
console.log("hi");
return undefined;
}
test().then(val => {
console.log(val); // undefined
});
A function that does not explexitly return something, actually returns undefined:
function test() { }
console.log(test());
Its the same with async functions, they also resolve to undefined when no other value was returned.
An async function doesn't need a return value in order to resolve. It is considered resolved when it finishes executing without an error.
If you throw new Error() inside of func2, console.log("main") will never be executed.
When you don't return from a function it implicitly returns a default value — normally undefined. So the promise returned by func2 will still resolve when the function returns.
From MDN:
A function without a return statement will return a default value. In
the case of a constructor called with the new keyword, the default
value is the value of its this parameter. For all other functions, the
default return value is undefined.
You can see this in your code if you alter it to this:
(async () => {
func2()
.then(d => console.log(d));
})()
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 read that async functions marked by the async keyword implicitly return a promise:
async function getVal(){
return await doSomethingAync();
}
var ret = getVal();
console.log(ret);
but that is not coherent...assuming doSomethingAsync() returns a promise, and the await keyword will return the value from the promise, not the promise itsef, then my getVal function should return that value, not an implicit promise.
So what exactly is the case? Do functions marked by the async keyword implicitly return promises or do we control what they return?
Perhaps if we don't explicitly return something, then they implicitly return a promise...?
To be more clear, there is a difference between the above and
function doSomethingAync(charlie) {
return new Promise(function (resolve) {
setTimeout(function () {
resolve(charlie || 'yikes');
}, 100);
})
}
async function getVal(){
var val = await doSomethingAync(); // val is not a promise
console.log(val); // logs 'yikes' or whatever
return val; // but this returns a promise
}
var ret = getVal();
console.log(ret); //logs a promise
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.
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);
Yes, an async function will always return a promise.
According to the tc39 spec, an async function desugars to a generator which yields Promises.
Specifically:
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.
Just add await before your function when you call it :
var ret = await getVal();
console.log(ret);
async doesn't return the promise, the await keyword awaits the resolution of the promise. async is an enhanced generator function and await works a bit like yield
I think the syntax (I am not 100% sure) is
async function* getVal() {...}
ES2016 generator functions work a bit like this. I have made a database handler based in top of tedious which you program like this
db.exec(function*(connection) {
if (params.passwd1 === '') {
let sql = 'UPDATE People SET UserName = #username WHERE ClinicianID = #clinicianid';
let request = connection.request(sql);
request.addParameter('username',db.TYPES.VarChar,params.username);
request.addParameter('clinicianid',db.TYPES.Int,uid);
yield connection.execSql();
} else {
if (!/^\S{4,}$/.test(params.passwd1)) {
response.end(JSON.stringify(
{status: false, passwd1: false,passwd2: true}
));
return;
}
let request = connection.request('SetPassword');
request.addParameter('userID',db.TYPES.Int,uid);
request.addParameter('username',db.TYPES.NVarChar,params.username);
request.addParameter('password',db.TYPES.VarChar,params.passwd1);
yield connection.callProcedure();
}
response.end(JSON.stringify({status: true}));
}).catch(err => {
logger('database',err.message);
response.end(JSON.stringify({status: false,passwd1: false,passwd2: false}));
});
Notice how I just program it like normal synchronous particularly at
yield connection.execSql and at yield connection.callProcedure
The db.exec function is a fairly typical Promise based generator
exec(generator) {
var self = this;
var it;
return new Promise((accept,reject) => {
var myConnection;
var onResult = lastPromiseResult => {
var obj = it.next(lastPromiseResult);
if (!obj.done) {
obj.value.then(onResult,reject);
} else {
if (myConnection) {
myConnection.release();
}
accept(obj.value);
}
};
self._connection().then(connection => {
myConnection = connection;
it = generator(connection); //This passes it into the generator
onResult(); //starts the generator
}).catch(error => {
reject(error);
});
});
}
I read that async functions marked by the async keyword implicitly return a promise:
async function getVal(){
return await doSomethingAync();
}
var ret = getVal();
console.log(ret);
but that is not coherent...assuming doSomethingAsync() returns a promise, and the await keyword will return the value from the promise, not the promise itsef, then my getVal function should return that value, not an implicit promise.
So what exactly is the case? Do functions marked by the async keyword implicitly return promises or do we control what they return?
Perhaps if we don't explicitly return something, then they implicitly return a promise...?
To be more clear, there is a difference between the above and
function doSomethingAync(charlie) {
return new Promise(function (resolve) {
setTimeout(function () {
resolve(charlie || 'yikes');
}, 100);
})
}
async function getVal(){
var val = await doSomethingAync(); // val is not a promise
console.log(val); // logs 'yikes' or whatever
return val; // but this returns a promise
}
var ret = getVal();
console.log(ret); //logs a promise
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.
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);
Yes, an async function will always return a promise.
According to the tc39 spec, an async function desugars to a generator which yields Promises.
Specifically:
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.
Just add await before your function when you call it :
var ret = await getVal();
console.log(ret);
async doesn't return the promise, the await keyword awaits the resolution of the promise. async is an enhanced generator function and await works a bit like yield
I think the syntax (I am not 100% sure) is
async function* getVal() {...}
ES2016 generator functions work a bit like this. I have made a database handler based in top of tedious which you program like this
db.exec(function*(connection) {
if (params.passwd1 === '') {
let sql = 'UPDATE People SET UserName = #username WHERE ClinicianID = #clinicianid';
let request = connection.request(sql);
request.addParameter('username',db.TYPES.VarChar,params.username);
request.addParameter('clinicianid',db.TYPES.Int,uid);
yield connection.execSql();
} else {
if (!/^\S{4,}$/.test(params.passwd1)) {
response.end(JSON.stringify(
{status: false, passwd1: false,passwd2: true}
));
return;
}
let request = connection.request('SetPassword');
request.addParameter('userID',db.TYPES.Int,uid);
request.addParameter('username',db.TYPES.NVarChar,params.username);
request.addParameter('password',db.TYPES.VarChar,params.passwd1);
yield connection.callProcedure();
}
response.end(JSON.stringify({status: true}));
}).catch(err => {
logger('database',err.message);
response.end(JSON.stringify({status: false,passwd1: false,passwd2: false}));
});
Notice how I just program it like normal synchronous particularly at
yield connection.execSql and at yield connection.callProcedure
The db.exec function is a fairly typical Promise based generator
exec(generator) {
var self = this;
var it;
return new Promise((accept,reject) => {
var myConnection;
var onResult = lastPromiseResult => {
var obj = it.next(lastPromiseResult);
if (!obj.done) {
obj.value.then(onResult,reject);
} else {
if (myConnection) {
myConnection.release();
}
accept(obj.value);
}
};
self._connection().then(connection => {
myConnection = connection;
it = generator(connection); //This passes it into the generator
onResult(); //starts the generator
}).catch(error => {
reject(error);
});
});
}
I read that async functions marked by the async keyword implicitly return a promise:
async function getVal(){
return await doSomethingAync();
}
var ret = getVal();
console.log(ret);
but that is not coherent...assuming doSomethingAsync() returns a promise, and the await keyword will return the value from the promise, not the promise itsef, then my getVal function should return that value, not an implicit promise.
So what exactly is the case? Do functions marked by the async keyword implicitly return promises or do we control what they return?
Perhaps if we don't explicitly return something, then they implicitly return a promise...?
To be more clear, there is a difference between the above and
function doSomethingAync(charlie) {
return new Promise(function (resolve) {
setTimeout(function () {
resolve(charlie || 'yikes');
}, 100);
})
}
async function getVal(){
var val = await doSomethingAync(); // val is not a promise
console.log(val); // logs 'yikes' or whatever
return val; // but this returns a promise
}
var ret = getVal();
console.log(ret); //logs a promise
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.
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);
Yes, an async function will always return a promise.
According to the tc39 spec, an async function desugars to a generator which yields Promises.
Specifically:
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.
Just add await before your function when you call it :
var ret = await getVal();
console.log(ret);
async doesn't return the promise, the await keyword awaits the resolution of the promise. async is an enhanced generator function and await works a bit like yield
I think the syntax (I am not 100% sure) is
async function* getVal() {...}
ES2016 generator functions work a bit like this. I have made a database handler based in top of tedious which you program like this
db.exec(function*(connection) {
if (params.passwd1 === '') {
let sql = 'UPDATE People SET UserName = #username WHERE ClinicianID = #clinicianid';
let request = connection.request(sql);
request.addParameter('username',db.TYPES.VarChar,params.username);
request.addParameter('clinicianid',db.TYPES.Int,uid);
yield connection.execSql();
} else {
if (!/^\S{4,}$/.test(params.passwd1)) {
response.end(JSON.stringify(
{status: false, passwd1: false,passwd2: true}
));
return;
}
let request = connection.request('SetPassword');
request.addParameter('userID',db.TYPES.Int,uid);
request.addParameter('username',db.TYPES.NVarChar,params.username);
request.addParameter('password',db.TYPES.VarChar,params.passwd1);
yield connection.callProcedure();
}
response.end(JSON.stringify({status: true}));
}).catch(err => {
logger('database',err.message);
response.end(JSON.stringify({status: false,passwd1: false,passwd2: false}));
});
Notice how I just program it like normal synchronous particularly at
yield connection.execSql and at yield connection.callProcedure
The db.exec function is a fairly typical Promise based generator
exec(generator) {
var self = this;
var it;
return new Promise((accept,reject) => {
var myConnection;
var onResult = lastPromiseResult => {
var obj = it.next(lastPromiseResult);
if (!obj.done) {
obj.value.then(onResult,reject);
} else {
if (myConnection) {
myConnection.release();
}
accept(obj.value);
}
};
self._connection().then(connection => {
myConnection = connection;
it = generator(connection); //This passes it into the generator
onResult(); //starts the generator
}).catch(error => {
reject(error);
});
});
}
Modern JS/Promise newbie here. I'm having trouble finding the answer to a simple question, despite the overwhelming amount of content on this subject.
I believe I have a decent understanding of JS Promises (thanks to various sources, including mdn and https://spin.atomicobject.com/2016/02/16/how-javascript-promises-work/)
I have consumed and produced Promises regularly in their simplest form, however I keep stumbling on methods in my inherited projects with the following pattern:
const doAsyncOp = () =>
$http.get(uri)
.then(
value => value,
reject => //...
);
My big question is: What happens when you simply return a value from your success handler? I need to consume this method, and need access to 'value' in the client code. Is this technically unhandled? Should I rewrite the implementation?
My big question is: What happens when you simply return a value from your success handler?
When you return a value from a .then() handler, that value becomes the resolved value of the parent promise chain: So, in this code:
// utility function that returns a promise that resolves after a delay
function delay(t, v) {
return new Promise(resolve => {
setTimeout(resolve.bind(null, v), t);
});
}
function test() {
return delay(100, 5).then(val => {
// this return value becomes the resolved value of the promise chain
// that is returned from this function
return 10;
});
}
test().then(result => {
// logs 10
console.log(result);
});
When you run it, it will log 10 because of the return 10 in the .then() handler.
There are four possibilities from a .then() handler:
Return a regular value such as return 10 or return val. That value becomes the resolved value of the promise chain. If no value is returned (which in Javascript means the return value is undefined), then the resolved value of the promise chain is undefined.
Return a promise that ultimately resolves or is already resolved. This promise is added to the chain and the promise chain takes on the resolved value of that promise.
Return a promise that ultimately rejects or is already rejected. This promise is added to the chain and the promise chain takes on the rejected reason of the returned promise.
Throw an exception. If an exception is thrown inside the .then() handler, then the .then() infrastructure will catch it and turn the promise chain into a rejected state with the reject reason set to the value that is thrown. So, if you do throw new Error("User not found") inside a .then() handler, then that promise chain will be rejected with that error object as the reject reason.
In your specific code:
const doAsyncOp = () =>
$http.get(uri)
.then(
value => value,
reject => //...
);
There is no reason for the value => value. value is already the resolved value of the promise, you do not need to set it again.
And since the fat arrow function automatically returns any single statement, you're already returning the promise from $http.get().then() in your doAsyncOp() function. So, you can just do:
const doAsyncOp = () => $http.get(uri);
doAsyncOp().then(result => {
// handle the result here
}).catch(err => {
// handle error here
});
To get this to the client code, just return the promise from your function:
const doAsyncOp = () => $http.get(uri)
then in your client you can use:
doAsyncOp()
.then(value => {
// use value
}
.catch(err => { /* error */ }
I think you can use async and await for the resolved values.
function delay(t, v) {
return new Promise(resolve => {
setTimeout(resolve.bind(null, v), t);
});
}
async function(){
await myData = delay(1000, 1000);
console.log(myData);
}