Why do the resolve message appear first? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Data science time! April 2019 and salary with experience Should we burninate the [wrap] tag? The Ask Question Wizard is Live!Why is using “for…in” with array iteration a bad idea?How do I make the first letter of a string uppercase in JavaScript?What is JSONP, and why was it created?Why does Google prepend while(1); to their JSON responses?Why does ++[[]][+[]]+[+[]] return the string “10”?What is TypeScript and why would I use it in place of JavaScript?ECMAScript 6 Chaining PromisesjQuery deferred.notify not workingjQuery Deferred continue to resolve after fail handlerCatching All Promise Rejections in an Async Function in JavaScript

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Why do the resolve message appear first?



Announcing the arrival of Valued Associate #679: Cesar Manara
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Data science time! April 2019 and salary with experience
Should we burninate the [wrap] tag?
The Ask Question Wizard is Live!Why is using “for…in” with array iteration a bad idea?How do I make the first letter of a string uppercase in JavaScript?What is JSONP, and why was it created?Why does Google prepend while(1); to their JSON responses?Why does ++[[]][+[]]+[+[]] return the string “10”?What is TypeScript and why would I use it in place of JavaScript?ECMAScript 6 Chaining PromisesjQuery deferred.notify not workingjQuery Deferred continue to resolve after fail handlerCatching All Promise Rejections in an Async Function in JavaScript



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9















I'm trying to wrap my mind around promises in Javascript. I was in the illusion that once a Promise was resolved it could never go back to rejected. To test that I wrote a little script. I see that the first messages that come back are the resolve messages "1 resolve 2" etc. I did expect the first message to be "0 reject 1".



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)

)

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)



at the console:



[Log] 1 resolve 2
[Log] 3 resolve 2
[Log] 5 resolve 2
[Log] 7 resolve 2
[Log] 9 resolve 2
[Log] 0 reject 1
[Log] 2 reject 1
[Log] 4 reject 1
[Log] 6 reject 1
[Log] 8 reject 1
< Promise status: "pending"


thanks for you help....



After reading



https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then



I got to this code. The catch is removed.



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)


)

p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)



at the console:



[Log] 0 reject 1
[Log] 1 resolve 2
[Log] 2 reject 1
[Log] 3 resolve 2
[Log] 4 reject 1
[Log] 5 resolve 2
[Log] 6 reject 1
[Log] 7 resolve 2
[Log] 8 reject 1
[Log] 9 resolve 2
< Promise status: "pending"









share|improve this question



















  • 3





    You are creating new promise on every iteration

    – brk
    1 hour ago











  • You have 10 promises, they don't "go back to rejected" like you said.

    – Kev
    1 hour ago






  • 3





    Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?

    – junvar
    1 hour ago






  • 2





    Yes Junvar. That is my question.

    – Edwin
    1 hour ago






  • 3





    99% sure it's because the .then and .catch each take a tick on the event loop. So the rejections are all a single tick behind your resolves.

    – jhpratt
    1 hour ago

















9















I'm trying to wrap my mind around promises in Javascript. I was in the illusion that once a Promise was resolved it could never go back to rejected. To test that I wrote a little script. I see that the first messages that come back are the resolve messages "1 resolve 2" etc. I did expect the first message to be "0 reject 1".



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)

)

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)



at the console:



[Log] 1 resolve 2
[Log] 3 resolve 2
[Log] 5 resolve 2
[Log] 7 resolve 2
[Log] 9 resolve 2
[Log] 0 reject 1
[Log] 2 reject 1
[Log] 4 reject 1
[Log] 6 reject 1
[Log] 8 reject 1
< Promise status: "pending"


thanks for you help....



After reading



https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then



I got to this code. The catch is removed.



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)


)

p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)



at the console:



[Log] 0 reject 1
[Log] 1 resolve 2
[Log] 2 reject 1
[Log] 3 resolve 2
[Log] 4 reject 1
[Log] 5 resolve 2
[Log] 6 reject 1
[Log] 7 resolve 2
[Log] 8 reject 1
[Log] 9 resolve 2
< Promise status: "pending"









share|improve this question



















  • 3





    You are creating new promise on every iteration

    – brk
    1 hour ago











  • You have 10 promises, they don't "go back to rejected" like you said.

    – Kev
    1 hour ago






  • 3





    Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?

    – junvar
    1 hour ago






  • 2





    Yes Junvar. That is my question.

    – Edwin
    1 hour ago






  • 3





    99% sure it's because the .then and .catch each take a tick on the event loop. So the rejections are all a single tick behind your resolves.

    – jhpratt
    1 hour ago













9












9








9


3






I'm trying to wrap my mind around promises in Javascript. I was in the illusion that once a Promise was resolved it could never go back to rejected. To test that I wrote a little script. I see that the first messages that come back are the resolve messages "1 resolve 2" etc. I did expect the first message to be "0 reject 1".



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)

)

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)



at the console:



[Log] 1 resolve 2
[Log] 3 resolve 2
[Log] 5 resolve 2
[Log] 7 resolve 2
[Log] 9 resolve 2
[Log] 0 reject 1
[Log] 2 reject 1
[Log] 4 reject 1
[Log] 6 reject 1
[Log] 8 reject 1
< Promise status: "pending"


thanks for you help....



After reading



https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then



I got to this code. The catch is removed.



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)


)

p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)



at the console:



[Log] 0 reject 1
[Log] 1 resolve 2
[Log] 2 reject 1
[Log] 3 resolve 2
[Log] 4 reject 1
[Log] 5 resolve 2
[Log] 6 reject 1
[Log] 7 resolve 2
[Log] 8 reject 1
[Log] 9 resolve 2
< Promise status: "pending"









share|improve this question
















I'm trying to wrap my mind around promises in Javascript. I was in the illusion that once a Promise was resolved it could never go back to rejected. To test that I wrote a little script. I see that the first messages that come back are the resolve messages "1 resolve 2" etc. I did expect the first message to be "0 reject 1".



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)

)

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)



at the console:



[Log] 1 resolve 2
[Log] 3 resolve 2
[Log] 5 resolve 2
[Log] 7 resolve 2
[Log] 9 resolve 2
[Log] 0 reject 1
[Log] 2 reject 1
[Log] 4 reject 1
[Log] 6 reject 1
[Log] 8 reject 1
< Promise status: "pending"


thanks for you help....



After reading



https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then



I got to this code. The catch is removed.



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)


)

p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)



at the console:



[Log] 0 reject 1
[Log] 1 resolve 2
[Log] 2 reject 1
[Log] 3 resolve 2
[Log] 4 reject 1
[Log] 5 resolve 2
[Log] 6 reject 1
[Log] 7 resolve 2
[Log] 8 reject 1
[Log] 9 resolve 2
< Promise status: "pending"






javascript es6-promise






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 38 mins ago







Edwin

















asked 1 hour ago









EdwinEdwin

513




513







  • 3





    You are creating new promise on every iteration

    – brk
    1 hour ago











  • You have 10 promises, they don't "go back to rejected" like you said.

    – Kev
    1 hour ago






  • 3





    Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?

    – junvar
    1 hour ago






  • 2





    Yes Junvar. That is my question.

    – Edwin
    1 hour ago






  • 3





    99% sure it's because the .then and .catch each take a tick on the event loop. So the rejections are all a single tick behind your resolves.

    – jhpratt
    1 hour ago












  • 3





    You are creating new promise on every iteration

    – brk
    1 hour ago











  • You have 10 promises, they don't "go back to rejected" like you said.

    – Kev
    1 hour ago






  • 3





    Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?

    – junvar
    1 hour ago






  • 2





    Yes Junvar. That is my question.

    – Edwin
    1 hour ago






  • 3





    99% sure it's because the .then and .catch each take a tick on the event loop. So the rejections are all a single tick behind your resolves.

    – jhpratt
    1 hour ago







3




3





You are creating new promise on every iteration

– brk
1 hour ago





You are creating new promise on every iteration

– brk
1 hour ago













You have 10 promises, they don't "go back to rejected" like you said.

– Kev
1 hour ago





You have 10 promises, they don't "go back to rejected" like you said.

– Kev
1 hour ago




3




3





Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?

– junvar
1 hour ago





Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?

– junvar
1 hour ago




2




2





Yes Junvar. That is my question.

– Edwin
1 hour ago





Yes Junvar. That is my question.

– Edwin
1 hour ago




3




3





99% sure it's because the .then and .catch each take a tick on the event loop. So the rejections are all a single tick behind your resolves.

– jhpratt
1 hour ago





99% sure it's because the .then and .catch each take a tick on the event loop. So the rejections are all a single tick behind your resolves.

– jhpratt
1 hour ago












4 Answers
4






active

oldest

votes


















3














You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):



enter image description here



As you can see, all your 10 promises are created before any of them are executed (resolve/reject).



Interestingly, in your code the resolved promises are handled first.



If you define the handlers in two separate definitions, you'll get the expected results:



p.then((message) => 
console.log(message)
)
p.catch((message) =>
console.log(message)
)


Output:



enter image description here






share|improve this answer























  • i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing

    – Yanis-git
    41 mins ago


















2














The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.



But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.



What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)

)

await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)



Or you could try use Promise.all() which will basically solve the order for you, see the official docs here






share|improve this answer


















  • 1





    well, you can remove the .then if you're using await

    – pushkin
    50 mins ago






  • 1





    Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.

    – Edwin
    47 mins ago


















0














Because JavaScript are mono thread :



  • promise

  • eventListener

  • setTimeout

  • setInterval

previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here



Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.



is Why Following code :






 let p = new Promise((resolve, reject) => 
resolve('micro task thread');
);

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');





will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.



Most of the time, event loop will execute the collection as first come first rendered.






share|improve this answer























  • @Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.

    – Yanis-git
    39 mins ago


















0














I did found a solution here:
MDN promise then



for (let i = 0; i < 10; i++) 

let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)


)

p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)







share|improve this answer























  • I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).

    – jhpratt
    30 mins ago











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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):



enter image description here



As you can see, all your 10 promises are created before any of them are executed (resolve/reject).



Interestingly, in your code the resolved promises are handled first.



If you define the handlers in two separate definitions, you'll get the expected results:



p.then((message) => 
console.log(message)
)
p.catch((message) =>
console.log(message)
)


Output:



enter image description here






share|improve this answer























  • i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing

    – Yanis-git
    41 mins ago















3














You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):



enter image description here



As you can see, all your 10 promises are created before any of them are executed (resolve/reject).



Interestingly, in your code the resolved promises are handled first.



If you define the handlers in two separate definitions, you'll get the expected results:



p.then((message) => 
console.log(message)
)
p.catch((message) =>
console.log(message)
)


Output:



enter image description here






share|improve this answer























  • i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing

    – Yanis-git
    41 mins ago













3












3








3







You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):



enter image description here



As you can see, all your 10 promises are created before any of them are executed (resolve/reject).



Interestingly, in your code the resolved promises are handled first.



If you define the handlers in two separate definitions, you'll get the expected results:



p.then((message) => 
console.log(message)
)
p.catch((message) =>
console.log(message)
)


Output:



enter image description here






share|improve this answer













You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):



enter image description here



As you can see, all your 10 promises are created before any of them are executed (resolve/reject).



Interestingly, in your code the resolved promises are handled first.



If you define the handlers in two separate definitions, you'll get the expected results:



p.then((message) => 
console.log(message)
)
p.catch((message) =>
console.log(message)
)


Output:



enter image description here







share|improve this answer












share|improve this answer



share|improve this answer










answered 1 hour ago









ludovicoludovico

775




775












  • i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing

    – Yanis-git
    41 mins ago

















  • i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing

    – Yanis-git
    41 mins ago
















i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing

– Yanis-git
41 mins ago





i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing

– Yanis-git
41 mins ago













2














The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.



But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.



What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)

)

await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)



Or you could try use Promise.all() which will basically solve the order for you, see the official docs here






share|improve this answer


















  • 1





    well, you can remove the .then if you're using await

    – pushkin
    50 mins ago






  • 1





    Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.

    – Edwin
    47 mins ago















2














The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.



But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.



What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)

)

await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)



Or you could try use Promise.all() which will basically solve the order for you, see the official docs here






share|improve this answer


















  • 1





    well, you can remove the .then if you're using await

    – pushkin
    50 mins ago






  • 1





    Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.

    – Edwin
    47 mins ago













2












2








2







The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.



But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.



What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)

)

await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)



Or you could try use Promise.all() which will basically solve the order for you, see the official docs here






share|improve this answer













The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.



But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.



What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:



for (let i = 0; i < 10; i++) 
let p = new Promise((resolve, reject) =>

let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)

)

await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)



Or you could try use Promise.all() which will basically solve the order for you, see the official docs here







share|improve this answer












share|improve this answer



share|improve this answer










answered 1 hour ago









Esdras XavierEsdras Xavier

45017




45017







  • 1





    well, you can remove the .then if you're using await

    – pushkin
    50 mins ago






  • 1





    Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.

    – Edwin
    47 mins ago












  • 1





    well, you can remove the .then if you're using await

    – pushkin
    50 mins ago






  • 1





    Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.

    – Edwin
    47 mins ago







1




1





well, you can remove the .then if you're using await

– pushkin
50 mins ago





well, you can remove the .then if you're using await

– pushkin
50 mins ago




1




1





Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.

– Edwin
47 mins ago





Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.

– Edwin
47 mins ago











0














Because JavaScript are mono thread :



  • promise

  • eventListener

  • setTimeout

  • setInterval

previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here



Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.



is Why Following code :






 let p = new Promise((resolve, reject) => 
resolve('micro task thread');
);

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');





will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.



Most of the time, event loop will execute the collection as first come first rendered.






share|improve this answer























  • @Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.

    – Yanis-git
    39 mins ago















0














Because JavaScript are mono thread :



  • promise

  • eventListener

  • setTimeout

  • setInterval

previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here



Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.



is Why Following code :






 let p = new Promise((resolve, reject) => 
resolve('micro task thread');
);

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');





will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.



Most of the time, event loop will execute the collection as first come first rendered.






share|improve this answer























  • @Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.

    – Yanis-git
    39 mins ago













0












0








0







Because JavaScript are mono thread :



  • promise

  • eventListener

  • setTimeout

  • setInterval

previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here



Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.



is Why Following code :






 let p = new Promise((resolve, reject) => 
resolve('micro task thread');
);

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');





will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.



Most of the time, event loop will execute the collection as first come first rendered.






share|improve this answer













Because JavaScript are mono thread :



  • promise

  • eventListener

  • setTimeout

  • setInterval

previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here



Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.



is Why Following code :






 let p = new Promise((resolve, reject) => 
resolve('micro task thread');
);

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');





will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.



Most of the time, event loop will execute the collection as first come first rendered.






 let p = new Promise((resolve, reject) => 
resolve('micro task thread');
);

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');





 let p = new Promise((resolve, reject) => 
resolve('micro task thread');
);

p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');






share|improve this answer












share|improve this answer



share|improve this answer










answered 49 mins ago









Yanis-gitYanis-git

2,6291725




2,6291725












  • @Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.

    – Yanis-git
    39 mins ago

















  • @Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.

    – Yanis-git
    39 mins ago
















@Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.

– Yanis-git
39 mins ago





@Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.

– Yanis-git
39 mins ago











0














I did found a solution here:
MDN promise then



for (let i = 0; i < 10; i++) 

let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)


)

p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)







share|improve this answer























  • I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).

    – jhpratt
    30 mins ago















0














I did found a solution here:
MDN promise then



for (let i = 0; i < 10; i++) 

let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)


)

p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)







share|improve this answer























  • I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).

    – jhpratt
    30 mins ago













0












0








0







I did found a solution here:
MDN promise then



for (let i = 0; i < 10; i++) 

let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)


)

p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)







share|improve this answer













I did found a solution here:
MDN promise then



for (let i = 0; i < 10; i++) 

let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)

if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)


)

p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)








share|improve this answer












share|improve this answer



share|improve this answer










answered 33 mins ago









EdwinEdwin

513




513












  • I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).

    – jhpratt
    30 mins ago

















  • I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).

    – jhpratt
    30 mins ago
















I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).

– jhpratt
30 mins ago





I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).

– jhpratt
30 mins ago

















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