Why does Async/Await work properly when the loop is inside the async function and not the other way around?












12















I have three snippets that loop three times while awaiting on a promise.



In the first snippet, it works as I expect and the value of i is decremented with each await.



let i = 3;

(async () => {
while (i) {
await Promise.resolve();
console.log(i);
i--;
}
})();


Output:



3
2
1




In the second one, the value of i is continuously decremented until it reaches zero and then all the awaits are executed.



let i = 3;

while (i) {
(async () => {
await Promise.resolve();
console.log(i);
})();
i--;
}


Output:



0
0
0




Lastly, this one causes an Allocation failed - JavaScript heap out of memory error and doesn't print any values.



let i = 3;
while (i) {
(async () => {
await Promise.resolve();
console.log(i);
i--;
})();
}




Can someone explain why they exhibit these different behaviors? Thanks.










share|improve this question




















  • 9





    Do you know how async functions desugar to promises?

    – Bergi
    yesterday











  • @Bergi It's just a syntactic sugar for dealing with promises instead of using the .then() and .catch() methods. Am I right ?

    – Ahmed Karaman
    10 hours ago











  • Yes, basically. So what do you think would your code look like if written with then()?

    – Bergi
    10 hours ago











  • For the second snippet, it would be: while (i) { Promise.resolve().then(res => console.log(i)); i--; }

    – Ahmed Karaman
    10 hours ago













  • Exactly. And it seems obvious now why this logs 0 three times.

    – Bergi
    8 hours ago
















12















I have three snippets that loop three times while awaiting on a promise.



In the first snippet, it works as I expect and the value of i is decremented with each await.



let i = 3;

(async () => {
while (i) {
await Promise.resolve();
console.log(i);
i--;
}
})();


Output:



3
2
1




In the second one, the value of i is continuously decremented until it reaches zero and then all the awaits are executed.



let i = 3;

while (i) {
(async () => {
await Promise.resolve();
console.log(i);
})();
i--;
}


Output:



0
0
0




Lastly, this one causes an Allocation failed - JavaScript heap out of memory error and doesn't print any values.



let i = 3;
while (i) {
(async () => {
await Promise.resolve();
console.log(i);
i--;
})();
}




Can someone explain why they exhibit these different behaviors? Thanks.










share|improve this question




















  • 9





    Do you know how async functions desugar to promises?

    – Bergi
    yesterday











  • @Bergi It's just a syntactic sugar for dealing with promises instead of using the .then() and .catch() methods. Am I right ?

    – Ahmed Karaman
    10 hours ago











  • Yes, basically. So what do you think would your code look like if written with then()?

    – Bergi
    10 hours ago











  • For the second snippet, it would be: while (i) { Promise.resolve().then(res => console.log(i)); i--; }

    – Ahmed Karaman
    10 hours ago













  • Exactly. And it seems obvious now why this logs 0 three times.

    – Bergi
    8 hours ago














12












12








12


1






I have three snippets that loop three times while awaiting on a promise.



In the first snippet, it works as I expect and the value of i is decremented with each await.



let i = 3;

(async () => {
while (i) {
await Promise.resolve();
console.log(i);
i--;
}
})();


Output:



3
2
1




In the second one, the value of i is continuously decremented until it reaches zero and then all the awaits are executed.



let i = 3;

while (i) {
(async () => {
await Promise.resolve();
console.log(i);
})();
i--;
}


Output:



0
0
0




Lastly, this one causes an Allocation failed - JavaScript heap out of memory error and doesn't print any values.



let i = 3;
while (i) {
(async () => {
await Promise.resolve();
console.log(i);
i--;
})();
}




Can someone explain why they exhibit these different behaviors? Thanks.










share|improve this question
















I have three snippets that loop three times while awaiting on a promise.



In the first snippet, it works as I expect and the value of i is decremented with each await.



let i = 3;

(async () => {
while (i) {
await Promise.resolve();
console.log(i);
i--;
}
})();


Output:



3
2
1




In the second one, the value of i is continuously decremented until it reaches zero and then all the awaits are executed.



let i = 3;

while (i) {
(async () => {
await Promise.resolve();
console.log(i);
})();
i--;
}


Output:



0
0
0




Lastly, this one causes an Allocation failed - JavaScript heap out of memory error and doesn't print any values.



let i = 3;
while (i) {
(async () => {
await Promise.resolve();
console.log(i);
i--;
})();
}




Can someone explain why they exhibit these different behaviors? Thanks.







javascript async-await es6-promise






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday







Ahmed Karaman

















asked yesterday









Ahmed KaramanAhmed Karaman

34719




34719








  • 9





    Do you know how async functions desugar to promises?

    – Bergi
    yesterday











  • @Bergi It's just a syntactic sugar for dealing with promises instead of using the .then() and .catch() methods. Am I right ?

    – Ahmed Karaman
    10 hours ago











  • Yes, basically. So what do you think would your code look like if written with then()?

    – Bergi
    10 hours ago











  • For the second snippet, it would be: while (i) { Promise.resolve().then(res => console.log(i)); i--; }

    – Ahmed Karaman
    10 hours ago













  • Exactly. And it seems obvious now why this logs 0 three times.

    – Bergi
    8 hours ago














  • 9





    Do you know how async functions desugar to promises?

    – Bergi
    yesterday











  • @Bergi It's just a syntactic sugar for dealing with promises instead of using the .then() and .catch() methods. Am I right ?

    – Ahmed Karaman
    10 hours ago











  • Yes, basically. So what do you think would your code look like if written with then()?

    – Bergi
    10 hours ago











  • For the second snippet, it would be: while (i) { Promise.resolve().then(res => console.log(i)); i--; }

    – Ahmed Karaman
    10 hours ago













  • Exactly. And it seems obvious now why this logs 0 three times.

    – Bergi
    8 hours ago








9




9





Do you know how async functions desugar to promises?

– Bergi
yesterday





Do you know how async functions desugar to promises?

– Bergi
yesterday













@Bergi It's just a syntactic sugar for dealing with promises instead of using the .then() and .catch() methods. Am I right ?

– Ahmed Karaman
10 hours ago





@Bergi It's just a syntactic sugar for dealing with promises instead of using the .then() and .catch() methods. Am I right ?

– Ahmed Karaman
10 hours ago













Yes, basically. So what do you think would your code look like if written with then()?

– Bergi
10 hours ago





Yes, basically. So what do you think would your code look like if written with then()?

– Bergi
10 hours ago













For the second snippet, it would be: while (i) { Promise.resolve().then(res => console.log(i)); i--; }

– Ahmed Karaman
10 hours ago







For the second snippet, it would be: while (i) { Promise.resolve().then(res => console.log(i)); i--; }

– Ahmed Karaman
10 hours ago















Exactly. And it seems obvious now why this logs 0 three times.

– Bergi
8 hours ago





Exactly. And it seems obvious now why this logs 0 three times.

– Bergi
8 hours ago












4 Answers
4






active

oldest

votes


















14














Concerning your second snippet:



Caling an async function without awaiting it's result is called fire and forget. You tell JavaScript that it should start some asynchronous processing, but you do not care when and how it finishes. That's what happens. It loops, fires some asynchronous tasks, when the loop is done they somewhen finish, and will log 0 as the loop already reached its end. If you'd do:



await (async () => {
await Promise.resolve();
console.log(i);
})();


it will loop in order.



Concerning your third snippet:



You never decrement i in the loop, therefore the loop runs forever. It would decrement i if the asynchronous tasks where executed somewhen, but that does not happen as the while loop runs crazy and blocks & crashes the browser.



 let i = 3;
while(i > 0) {
doStuff();
}





share|improve this answer

































    9














    Focusing primarily on the last example:



    let i = 3;
    while (i) {
    (async () => {
    await Promise.resolve();
    console.log(i);
    i--;
    })();
    }


    It may help if we rewrite the code without async/await to reveal what it is really doing. Under the hood, the code execution of the async function is deferred for later:



    let callbacks = ;

    let i = 0;
    while (i > 0) {
    callbacks.push(() => {
    console.log(i);
    i--;
    });
    }

    callbacks.forEach(cb => {
    cb();
    });


    As you can see, none of the callbacks are executed until after the loop is completed. Since the loop never halts, eventually the vm will run out of space to store callbacks.






    share|improve this answer































      2














      Because in the first case console.log and decrement work in sync with each other because they are both inside the same asnychronous function.
      In the second case console.log works asynchronously and decrement works synchronously.
      Therefore, the decrement will be executed first, the asynchronous function will wait for the synchronous function to finish, then it will be executed with i == 0



      In the third case, the loop body executes synchronously, and runs the asynchronous function at each iteration. Therefore, the decrement can not work until the end of the loop, so the condition in the cycle is always true. And so until the stack or memory is full






      share|improve this answer










      New contributor




      Nikita Umnov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




























        2














        In your particular example it decrements the i and then runs the async code like:



        let i = 3;

        while (i) {
        i--; // <---------------------
        (async () => { // |
        await Promise.resolve();// |
        console.log(i); // |
        })(); // |
        // >---------------------------
        }


        Regarding your third snippet, it will never decrease the i value and so loop runs forever and thus crashes application:



        let i = 3;
        while (i) {
        (async () => {
        await Promise.resolve(); // await and resolve >-----------
        // the following code doesn't run after it resolves // |
        console.log(i); // |
        i--; // |
        })(); // |
        // out from the (async() => {})() <-------------------------
        }





        share|improve this answer


























        • Not really, If you would remove the await Promise.resolve it would log correctly.

          – Jonas Wilms
          yesterday











        • @Bhojendra Rauniyar Can you please check this third snippet?

          – Ahmed Karaman
          yesterday











        • @JonasWilms Does what you say mean that if we have an async function without an await in its body that it will be blocking and run synchronously like any other function?

          – Ahmed Karaman
          yesterday








        • 2





          the asynchronous code runs in background this is incorrect. The async code is deferred, and since event loop is busy with infinite while loop the async code never runs. if it was in background, the output numbers would be random but the i would be equal to zero eventually.

          – meze
          yesterday






        • 1





          @BhojendraRauniyar JavaScript is single-threaded, so there is no “background”. Only one chunk of code can run at one time. I’m fuzzing over some of the details here, but there is a single event loop that dequeues a pending callback and runs it to completion. In order to prevent blocking (such as waiting for a timer or a network request), code is broken up into smaller callbacks in order to give competing events their chance to run in the single thread. await is just syntactic sugar for “schedule a callback to run after the thing we’re waiting on is available”.

          – Joel Cornett
          yesterday











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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        14














        Concerning your second snippet:



        Caling an async function without awaiting it's result is called fire and forget. You tell JavaScript that it should start some asynchronous processing, but you do not care when and how it finishes. That's what happens. It loops, fires some asynchronous tasks, when the loop is done they somewhen finish, and will log 0 as the loop already reached its end. If you'd do:



        await (async () => {
        await Promise.resolve();
        console.log(i);
        })();


        it will loop in order.



        Concerning your third snippet:



        You never decrement i in the loop, therefore the loop runs forever. It would decrement i if the asynchronous tasks where executed somewhen, but that does not happen as the while loop runs crazy and blocks & crashes the browser.



         let i = 3;
        while(i > 0) {
        doStuff();
        }





        share|improve this answer






























          14














          Concerning your second snippet:



          Caling an async function without awaiting it's result is called fire and forget. You tell JavaScript that it should start some asynchronous processing, but you do not care when and how it finishes. That's what happens. It loops, fires some asynchronous tasks, when the loop is done they somewhen finish, and will log 0 as the loop already reached its end. If you'd do:



          await (async () => {
          await Promise.resolve();
          console.log(i);
          })();


          it will loop in order.



          Concerning your third snippet:



          You never decrement i in the loop, therefore the loop runs forever. It would decrement i if the asynchronous tasks where executed somewhen, but that does not happen as the while loop runs crazy and blocks & crashes the browser.



           let i = 3;
          while(i > 0) {
          doStuff();
          }





          share|improve this answer




























            14












            14








            14







            Concerning your second snippet:



            Caling an async function without awaiting it's result is called fire and forget. You tell JavaScript that it should start some asynchronous processing, but you do not care when and how it finishes. That's what happens. It loops, fires some asynchronous tasks, when the loop is done they somewhen finish, and will log 0 as the loop already reached its end. If you'd do:



            await (async () => {
            await Promise.resolve();
            console.log(i);
            })();


            it will loop in order.



            Concerning your third snippet:



            You never decrement i in the loop, therefore the loop runs forever. It would decrement i if the asynchronous tasks where executed somewhen, but that does not happen as the while loop runs crazy and blocks & crashes the browser.



             let i = 3;
            while(i > 0) {
            doStuff();
            }





            share|improve this answer















            Concerning your second snippet:



            Caling an async function without awaiting it's result is called fire and forget. You tell JavaScript that it should start some asynchronous processing, but you do not care when and how it finishes. That's what happens. It loops, fires some asynchronous tasks, when the loop is done they somewhen finish, and will log 0 as the loop already reached its end. If you'd do:



            await (async () => {
            await Promise.resolve();
            console.log(i);
            })();


            it will loop in order.



            Concerning your third snippet:



            You never decrement i in the loop, therefore the loop runs forever. It would decrement i if the asynchronous tasks where executed somewhen, but that does not happen as the while loop runs crazy and blocks & crashes the browser.



             let i = 3;
            while(i > 0) {
            doStuff();
            }






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 14 hours ago

























            answered yesterday









            Jonas WilmsJonas Wilms

            63.3k53457




            63.3k53457

























                9














                Focusing primarily on the last example:



                let i = 3;
                while (i) {
                (async () => {
                await Promise.resolve();
                console.log(i);
                i--;
                })();
                }


                It may help if we rewrite the code without async/await to reveal what it is really doing. Under the hood, the code execution of the async function is deferred for later:



                let callbacks = ;

                let i = 0;
                while (i > 0) {
                callbacks.push(() => {
                console.log(i);
                i--;
                });
                }

                callbacks.forEach(cb => {
                cb();
                });


                As you can see, none of the callbacks are executed until after the loop is completed. Since the loop never halts, eventually the vm will run out of space to store callbacks.






                share|improve this answer




























                  9














                  Focusing primarily on the last example:



                  let i = 3;
                  while (i) {
                  (async () => {
                  await Promise.resolve();
                  console.log(i);
                  i--;
                  })();
                  }


                  It may help if we rewrite the code without async/await to reveal what it is really doing. Under the hood, the code execution of the async function is deferred for later:



                  let callbacks = ;

                  let i = 0;
                  while (i > 0) {
                  callbacks.push(() => {
                  console.log(i);
                  i--;
                  });
                  }

                  callbacks.forEach(cb => {
                  cb();
                  });


                  As you can see, none of the callbacks are executed until after the loop is completed. Since the loop never halts, eventually the vm will run out of space to store callbacks.






                  share|improve this answer


























                    9












                    9








                    9







                    Focusing primarily on the last example:



                    let i = 3;
                    while (i) {
                    (async () => {
                    await Promise.resolve();
                    console.log(i);
                    i--;
                    })();
                    }


                    It may help if we rewrite the code without async/await to reveal what it is really doing. Under the hood, the code execution of the async function is deferred for later:



                    let callbacks = ;

                    let i = 0;
                    while (i > 0) {
                    callbacks.push(() => {
                    console.log(i);
                    i--;
                    });
                    }

                    callbacks.forEach(cb => {
                    cb();
                    });


                    As you can see, none of the callbacks are executed until after the loop is completed. Since the loop never halts, eventually the vm will run out of space to store callbacks.






                    share|improve this answer













                    Focusing primarily on the last example:



                    let i = 3;
                    while (i) {
                    (async () => {
                    await Promise.resolve();
                    console.log(i);
                    i--;
                    })();
                    }


                    It may help if we rewrite the code without async/await to reveal what it is really doing. Under the hood, the code execution of the async function is deferred for later:



                    let callbacks = ;

                    let i = 0;
                    while (i > 0) {
                    callbacks.push(() => {
                    console.log(i);
                    i--;
                    });
                    }

                    callbacks.forEach(cb => {
                    cb();
                    });


                    As you can see, none of the callbacks are executed until after the loop is completed. Since the loop never halts, eventually the vm will run out of space to store callbacks.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered yesterday









                    Joel CornettJoel Cornett

                    17.9k44270




                    17.9k44270























                        2














                        Because in the first case console.log and decrement work in sync with each other because they are both inside the same asnychronous function.
                        In the second case console.log works asynchronously and decrement works synchronously.
                        Therefore, the decrement will be executed first, the asynchronous function will wait for the synchronous function to finish, then it will be executed with i == 0



                        In the third case, the loop body executes synchronously, and runs the asynchronous function at each iteration. Therefore, the decrement can not work until the end of the loop, so the condition in the cycle is always true. And so until the stack or memory is full






                        share|improve this answer










                        New contributor




                        Nikita Umnov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.

























                          2














                          Because in the first case console.log and decrement work in sync with each other because they are both inside the same asnychronous function.
                          In the second case console.log works asynchronously and decrement works synchronously.
                          Therefore, the decrement will be executed first, the asynchronous function will wait for the synchronous function to finish, then it will be executed with i == 0



                          In the third case, the loop body executes synchronously, and runs the asynchronous function at each iteration. Therefore, the decrement can not work until the end of the loop, so the condition in the cycle is always true. And so until the stack or memory is full






                          share|improve this answer










                          New contributor




                          Nikita Umnov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.























                            2












                            2








                            2







                            Because in the first case console.log and decrement work in sync with each other because they are both inside the same asnychronous function.
                            In the second case console.log works asynchronously and decrement works synchronously.
                            Therefore, the decrement will be executed first, the asynchronous function will wait for the synchronous function to finish, then it will be executed with i == 0



                            In the third case, the loop body executes synchronously, and runs the asynchronous function at each iteration. Therefore, the decrement can not work until the end of the loop, so the condition in the cycle is always true. And so until the stack or memory is full






                            share|improve this answer










                            New contributor




                            Nikita Umnov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.










                            Because in the first case console.log and decrement work in sync with each other because they are both inside the same asnychronous function.
                            In the second case console.log works asynchronously and decrement works synchronously.
                            Therefore, the decrement will be executed first, the asynchronous function will wait for the synchronous function to finish, then it will be executed with i == 0



                            In the third case, the loop body executes synchronously, and runs the asynchronous function at each iteration. Therefore, the decrement can not work until the end of the loop, so the condition in the cycle is always true. And so until the stack or memory is full







                            share|improve this answer










                            New contributor




                            Nikita Umnov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|improve this answer



                            share|improve this answer








                            edited yesterday





















                            New contributor




                            Nikita Umnov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            answered yesterday









                            Nikita UmnovNikita Umnov

                            1544




                            1544




                            New contributor




                            Nikita Umnov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.





                            New contributor





                            Nikita Umnov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            Nikita Umnov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.























                                2














                                In your particular example it decrements the i and then runs the async code like:



                                let i = 3;

                                while (i) {
                                i--; // <---------------------
                                (async () => { // |
                                await Promise.resolve();// |
                                console.log(i); // |
                                })(); // |
                                // >---------------------------
                                }


                                Regarding your third snippet, it will never decrease the i value and so loop runs forever and thus crashes application:



                                let i = 3;
                                while (i) {
                                (async () => {
                                await Promise.resolve(); // await and resolve >-----------
                                // the following code doesn't run after it resolves // |
                                console.log(i); // |
                                i--; // |
                                })(); // |
                                // out from the (async() => {})() <-------------------------
                                }





                                share|improve this answer


























                                • Not really, If you would remove the await Promise.resolve it would log correctly.

                                  – Jonas Wilms
                                  yesterday











                                • @Bhojendra Rauniyar Can you please check this third snippet?

                                  – Ahmed Karaman
                                  yesterday











                                • @JonasWilms Does what you say mean that if we have an async function without an await in its body that it will be blocking and run synchronously like any other function?

                                  – Ahmed Karaman
                                  yesterday








                                • 2





                                  the asynchronous code runs in background this is incorrect. The async code is deferred, and since event loop is busy with infinite while loop the async code never runs. if it was in background, the output numbers would be random but the i would be equal to zero eventually.

                                  – meze
                                  yesterday






                                • 1





                                  @BhojendraRauniyar JavaScript is single-threaded, so there is no “background”. Only one chunk of code can run at one time. I’m fuzzing over some of the details here, but there is a single event loop that dequeues a pending callback and runs it to completion. In order to prevent blocking (such as waiting for a timer or a network request), code is broken up into smaller callbacks in order to give competing events their chance to run in the single thread. await is just syntactic sugar for “schedule a callback to run after the thing we’re waiting on is available”.

                                  – Joel Cornett
                                  yesterday
















                                2














                                In your particular example it decrements the i and then runs the async code like:



                                let i = 3;

                                while (i) {
                                i--; // <---------------------
                                (async () => { // |
                                await Promise.resolve();// |
                                console.log(i); // |
                                })(); // |
                                // >---------------------------
                                }


                                Regarding your third snippet, it will never decrease the i value and so loop runs forever and thus crashes application:



                                let i = 3;
                                while (i) {
                                (async () => {
                                await Promise.resolve(); // await and resolve >-----------
                                // the following code doesn't run after it resolves // |
                                console.log(i); // |
                                i--; // |
                                })(); // |
                                // out from the (async() => {})() <-------------------------
                                }





                                share|improve this answer


























                                • Not really, If you would remove the await Promise.resolve it would log correctly.

                                  – Jonas Wilms
                                  yesterday











                                • @Bhojendra Rauniyar Can you please check this third snippet?

                                  – Ahmed Karaman
                                  yesterday











                                • @JonasWilms Does what you say mean that if we have an async function without an await in its body that it will be blocking and run synchronously like any other function?

                                  – Ahmed Karaman
                                  yesterday








                                • 2





                                  the asynchronous code runs in background this is incorrect. The async code is deferred, and since event loop is busy with infinite while loop the async code never runs. if it was in background, the output numbers would be random but the i would be equal to zero eventually.

                                  – meze
                                  yesterday






                                • 1





                                  @BhojendraRauniyar JavaScript is single-threaded, so there is no “background”. Only one chunk of code can run at one time. I’m fuzzing over some of the details here, but there is a single event loop that dequeues a pending callback and runs it to completion. In order to prevent blocking (such as waiting for a timer or a network request), code is broken up into smaller callbacks in order to give competing events their chance to run in the single thread. await is just syntactic sugar for “schedule a callback to run after the thing we’re waiting on is available”.

                                  – Joel Cornett
                                  yesterday














                                2












                                2








                                2







                                In your particular example it decrements the i and then runs the async code like:



                                let i = 3;

                                while (i) {
                                i--; // <---------------------
                                (async () => { // |
                                await Promise.resolve();// |
                                console.log(i); // |
                                })(); // |
                                // >---------------------------
                                }


                                Regarding your third snippet, it will never decrease the i value and so loop runs forever and thus crashes application:



                                let i = 3;
                                while (i) {
                                (async () => {
                                await Promise.resolve(); // await and resolve >-----------
                                // the following code doesn't run after it resolves // |
                                console.log(i); // |
                                i--; // |
                                })(); // |
                                // out from the (async() => {})() <-------------------------
                                }





                                share|improve this answer















                                In your particular example it decrements the i and then runs the async code like:



                                let i = 3;

                                while (i) {
                                i--; // <---------------------
                                (async () => { // |
                                await Promise.resolve();// |
                                console.log(i); // |
                                })(); // |
                                // >---------------------------
                                }


                                Regarding your third snippet, it will never decrease the i value and so loop runs forever and thus crashes application:



                                let i = 3;
                                while (i) {
                                (async () => {
                                await Promise.resolve(); // await and resolve >-----------
                                // the following code doesn't run after it resolves // |
                                console.log(i); // |
                                i--; // |
                                })(); // |
                                // out from the (async() => {})() <-------------------------
                                }






                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited yesterday

























                                answered yesterday









                                Bhojendra RauniyarBhojendra Rauniyar

                                53.1k2081135




                                53.1k2081135













                                • Not really, If you would remove the await Promise.resolve it would log correctly.

                                  – Jonas Wilms
                                  yesterday











                                • @Bhojendra Rauniyar Can you please check this third snippet?

                                  – Ahmed Karaman
                                  yesterday











                                • @JonasWilms Does what you say mean that if we have an async function without an await in its body that it will be blocking and run synchronously like any other function?

                                  – Ahmed Karaman
                                  yesterday








                                • 2





                                  the asynchronous code runs in background this is incorrect. The async code is deferred, and since event loop is busy with infinite while loop the async code never runs. if it was in background, the output numbers would be random but the i would be equal to zero eventually.

                                  – meze
                                  yesterday






                                • 1





                                  @BhojendraRauniyar JavaScript is single-threaded, so there is no “background”. Only one chunk of code can run at one time. I’m fuzzing over some of the details here, but there is a single event loop that dequeues a pending callback and runs it to completion. In order to prevent blocking (such as waiting for a timer or a network request), code is broken up into smaller callbacks in order to give competing events their chance to run in the single thread. await is just syntactic sugar for “schedule a callback to run after the thing we’re waiting on is available”.

                                  – Joel Cornett
                                  yesterday



















                                • Not really, If you would remove the await Promise.resolve it would log correctly.

                                  – Jonas Wilms
                                  yesterday











                                • @Bhojendra Rauniyar Can you please check this third snippet?

                                  – Ahmed Karaman
                                  yesterday











                                • @JonasWilms Does what you say mean that if we have an async function without an await in its body that it will be blocking and run synchronously like any other function?

                                  – Ahmed Karaman
                                  yesterday








                                • 2





                                  the asynchronous code runs in background this is incorrect. The async code is deferred, and since event loop is busy with infinite while loop the async code never runs. if it was in background, the output numbers would be random but the i would be equal to zero eventually.

                                  – meze
                                  yesterday






                                • 1





                                  @BhojendraRauniyar JavaScript is single-threaded, so there is no “background”. Only one chunk of code can run at one time. I’m fuzzing over some of the details here, but there is a single event loop that dequeues a pending callback and runs it to completion. In order to prevent blocking (such as waiting for a timer or a network request), code is broken up into smaller callbacks in order to give competing events their chance to run in the single thread. await is just syntactic sugar for “schedule a callback to run after the thing we’re waiting on is available”.

                                  – Joel Cornett
                                  yesterday

















                                Not really, If you would remove the await Promise.resolve it would log correctly.

                                – Jonas Wilms
                                yesterday





                                Not really, If you would remove the await Promise.resolve it would log correctly.

                                – Jonas Wilms
                                yesterday













                                @Bhojendra Rauniyar Can you please check this third snippet?

                                – Ahmed Karaman
                                yesterday





                                @Bhojendra Rauniyar Can you please check this third snippet?

                                – Ahmed Karaman
                                yesterday













                                @JonasWilms Does what you say mean that if we have an async function without an await in its body that it will be blocking and run synchronously like any other function?

                                – Ahmed Karaman
                                yesterday







                                @JonasWilms Does what you say mean that if we have an async function without an await in its body that it will be blocking and run synchronously like any other function?

                                – Ahmed Karaman
                                yesterday






                                2




                                2





                                the asynchronous code runs in background this is incorrect. The async code is deferred, and since event loop is busy with infinite while loop the async code never runs. if it was in background, the output numbers would be random but the i would be equal to zero eventually.

                                – meze
                                yesterday





                                the asynchronous code runs in background this is incorrect. The async code is deferred, and since event loop is busy with infinite while loop the async code never runs. if it was in background, the output numbers would be random but the i would be equal to zero eventually.

                                – meze
                                yesterday




                                1




                                1





                                @BhojendraRauniyar JavaScript is single-threaded, so there is no “background”. Only one chunk of code can run at one time. I’m fuzzing over some of the details here, but there is a single event loop that dequeues a pending callback and runs it to completion. In order to prevent blocking (such as waiting for a timer or a network request), code is broken up into smaller callbacks in order to give competing events their chance to run in the single thread. await is just syntactic sugar for “schedule a callback to run after the thing we’re waiting on is available”.

                                – Joel Cornett
                                yesterday





                                @BhojendraRauniyar JavaScript is single-threaded, so there is no “background”. Only one chunk of code can run at one time. I’m fuzzing over some of the details here, but there is a single event loop that dequeues a pending callback and runs it to completion. In order to prevent blocking (such as waiting for a timer or a network request), code is broken up into smaller callbacks in order to give competing events their chance to run in the single thread. await is just syntactic sugar for “schedule a callback to run after the thing we’re waiting on is available”.

                                – Joel Cornett
                                yesterday


















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