Mixing
Would ES7 behave like the function was fully synchronous here?
up vote
3
down vote
favorite
The following case:
I have a wrapperfunction "InitializePage()" which is called when loading the page.
This one contains the functions a-d out of which a and b contain AJAX. It would look like this:
function wrapperFunction() {
a() //contains AJAX
b() //contains AJAX
c() //Synchronous
d() //Synchronous
}
Now, I want function b to start only if function a has finished already.
I know that I could easily do this by making the wrapperFunction "async", use "await" in front of function a and then return the promise from function a (using JQuery Ajax) like this:
function a() {
return $.post('somefile.php', {
//someCode
})
}
But I want to know one thing:
function a by itself should be treated by JS as SYNCHRONOUS code until it hits the JqueryAJAX inside the function, right?
Only when it hits the AJAX call, JS hands it off to the C++ API and continues to the next bit of code no matter whether the AJAX call has finished execution yet or not, right?
So lets assume, even though it would be unnecessarily hacky, I would do this:
async function a() {
await $.post('someFile.php', {
//some code
})
}
Since I synchronized the AJAX part of function a(), would this mean that at the level of wrapperFunction(), JS does NOT procede until function a() and all of its contents have finished execution?
javascript jquery ajax ecmascript-7
asked 3 hours ago
JSONBUG123
527
|
show 1 more comment
up vote
3
down vote
favorite
The following case:
I have a wrapperfunction "InitializePage()" which is called when loading the page.
This one contains the functions a-d out of which a and b contain AJAX. It would look like this:
function wrapperFunction() {
a() //contains AJAX
b() //contains AJAX
c() //Synchronous
d() //Synchronous
}
Now, I want function b to start only if function a has finished already.
I know that I could easily do this by making the wrapperFunction "async", use "await" in front of function a and then return the promise from function a (using JQuery Ajax) like this:
function a() {
return $.post('somefile.php', {
//someCode
})
}
But I want to know one thing:
function a by itself should be treated by JS as SYNCHRONOUS code until it hits the JqueryAJAX inside the function, right?
Only when it hits the AJAX call, JS hands it off to the C++ API and continues to the next bit of code no matter whether the AJAX call has finished execution yet or not, right?
So lets assume, even though it would be unnecessarily hacky, I would do this:
async function a() {
await $.post('someFile.php', {
//some code
})
}
Since I synchronized the AJAX part of function a(), would this mean that at the level of wrapperFunction(), JS does NOT procede until function a() and all of its contents have finished execution?
javascript jquery ajax ecmascript-7
asked 3 hours ago
JSONBUG123
527
1
It might help to look at theawaitcode and think of what it would look like when it was written with.then()
– Bergi
3 hours ago
You need to writeawait a()inwrapperFunction()to make it seem like it's synchronous. And then you need to declareasync function wrapperFunction()
– Barmar
3 hours ago
@Barmar I know this way, but out of curiosity, I wanted to know whether the way I described would work as well or not.
– JSONBUG123
3 hours ago
The simple answer is no. Declaring a functionasyncis just syntactic sugar for automatically returning a promise. It doesn't make it act synchronous. Theawaitkeyword is what makes a call to an async function seem like it's synchronous.
– Barmar
3 hours ago
but the keyword "async" doesn't make a function return promises? Instead, an async function enables the await keyword for use INSIDE the async function, which can await promises being returned from any function inside the function which was prefixed with "async" .
– JSONBUG123
3 hours ago
|
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The following case:
I have a wrapperfunction "InitializePage()" which is called when loading the page.
This one contains the functions a-d out of which a and b contain AJAX. It would look like this:
function wrapperFunction() {
a() //contains AJAX
b() //contains AJAX
c() //Synchronous
d() //Synchronous
}
Now, I want function b to start only if function a has finished already.
I know that I could easily do this by making the wrapperFunction "async", use "await" in front of function a and then return the promise from function a (using JQuery Ajax) like this:
function a() {
return $.post('somefile.php', {
//someCode
})
}
But I want to know one thing:
function a by itself should be treated by JS as SYNCHRONOUS code until it hits the JqueryAJAX inside the function, right?
Only when it hits the AJAX call, JS hands it off to the C++ API and continues to the next bit of code no matter whether the AJAX call has finished execution yet or not, right?
So lets assume, even though it would be unnecessarily hacky, I would do this:
async function a() {
await $.post('someFile.php', {
//some code
})
}
Since I synchronized the AJAX part of function a(), would this mean that at the level of wrapperFunction(), JS does NOT procede until function a() and all of its contents have finished execution?
javascript jquery ajax ecmascript-7
asked 3 hours ago
JSONBUG123
527
The following case:
I have a wrapperfunction "InitializePage()" which is called when loading the page.
This one contains the functions a-d out of which a and b contain AJAX. It would look like this:
function wrapperFunction() {
a() //contains AJAX
b() //contains AJAX
c() //Synchronous
d() //Synchronous
}
Now, I want function b to start only if function a has finished already.
I know that I could easily do this by making the wrapperFunction "async", use "await" in front of function a and then return the promise from function a (using JQuery Ajax) like this:
function a() {
return $.post('somefile.php', {
//someCode
})
}
But I want to know one thing:
function a by itself should be treated by JS as SYNCHRONOUS code until it hits the JqueryAJAX inside the function, right?
Only when it hits the AJAX call, JS hands it off to the C++ API and continues to the next bit of code no matter whether the AJAX call has finished execution yet or not, right?
So lets assume, even though it would be unnecessarily hacky, I would do this:
async function a() {
await $.post('someFile.php', {
//some code
})
}
Since I synchronized the AJAX part of function a(), would this mean that at the level of wrapperFunction(), JS does NOT procede until function a() and all of its contents have finished execution?
javascript jquery ajax ecmascript-7
javascript jquery ajax ecmascript-7
asked 3 hours ago
JSONBUG123
527
asked 3 hours ago
JSONBUG123
527
edited 3 hours ago
asked 3 hours ago
JSONBUG123
527
asked 3 hours ago
JSONBUG123
527
asked 3 hours ago
JSONBUG123
527
527
1
It might help to look at theawaitcode and think of what it would look like when it was written with.then()
– Bergi
3 hours ago
You need to writeawait a()inwrapperFunction()to make it seem like it's synchronous. And then you need to declareasync function wrapperFunction()
– Barmar
3 hours ago
@Barmar I know this way, but out of curiosity, I wanted to know whether the way I described would work as well or not.
– JSONBUG123
3 hours ago
The simple answer is no. Declaring a functionasyncis just syntactic sugar for automatically returning a promise. It doesn't make it act synchronous. Theawaitkeyword is what makes a call to an async function seem like it's synchronous.
– Barmar
3 hours ago
but the keyword "async" doesn't make a function return promises? Instead, an async function enables the await keyword for use INSIDE the async function, which can await promises being returned from any function inside the function which was prefixed with "async" .
– JSONBUG123
3 hours ago
|
show 1 more comment
1
It might help to look at theawaitcode and think of what it would look like when it was written with.then()
– Bergi
3 hours ago
You need to writeawait a()inwrapperFunction()to make it seem like it's synchronous. And then you need to declareasync function wrapperFunction()
– Barmar
3 hours ago
@Barmar I know this way, but out of curiosity, I wanted to know whether the way I described would work as well or not.
– JSONBUG123
3 hours ago
The simple answer is no. Declaring a functionasyncis just syntactic sugar for automatically returning a promise. It doesn't make it act synchronous. Theawaitkeyword is what makes a call to an async function seem like it's synchronous.
– Barmar
3 hours ago
but the keyword "async" doesn't make a function return promises? Instead, an async function enables the await keyword for use INSIDE the async function, which can await promises being returned from any function inside the function which was prefixed with "async" .
– JSONBUG123
3 hours ago
1
1
It might help to look at the
await code and think of what it would look like when it was written with .then()– Bergi
3 hours ago
It might help to look at the
await code and think of what it would look like when it was written with .then()– Bergi
3 hours ago
You need to write
await a() in wrapperFunction() to make it seem like it's synchronous. And then you need to declare async function wrapperFunction()– Barmar
3 hours ago
You need to write
await a() in wrapperFunction() to make it seem like it's synchronous. And then you need to declare async function wrapperFunction()– Barmar
3 hours ago
@Barmar I know this way, but out of curiosity, I wanted to know whether the way I described would work as well or not.
– JSONBUG123
3 hours ago
@Barmar I know this way, but out of curiosity, I wanted to know whether the way I described would work as well or not.
– JSONBUG123
3 hours ago
The simple answer is no. Declaring a function
async is just syntactic sugar for automatically returning a promise. It doesn't make it act synchronous. The await keyword is what makes a call to an async function seem like it's synchronous.– Barmar
3 hours ago
The simple answer is no. Declaring a function
async is just syntactic sugar for automatically returning a promise. It doesn't make it act synchronous. The await keyword is what makes a call to an async function seem like it's synchronous.– Barmar
3 hours ago
but the keyword "async" doesn't make a function return promises? Instead, an async function enables the await keyword for use INSIDE the async function, which can await promises being returned from any function inside the function which was prefixed with "async" .
– JSONBUG123
3 hours ago
but the keyword "async" doesn't make a function return promises? Instead, an async function enables the await keyword for use INSIDE the async function, which can await promises being returned from any function inside the function which was prefixed with "async" .
– JSONBUG123
3 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The async function declaration defines an asynchronous function, which returns an AsyncFunction object. An asynchronous function is a function which operates asynchronously via the event loop, using an implicit Promise to return its result. But the syntax and structure of your code using async functions is much more like using standard synchronous functions.
So as per your question.
async function a() {
await $.post('someFile.php', {
//some code
})
}
yes it makes it asynchronous. but it is always better to use await where we are calling this function something like below.
for more refrenece please take a look here.
function resolveAfter2Seconds() {
return new Promise(resolve => {
setTimeout(() => {
resolve('resolved');
}, 2000);
});
}
async function asyncCall() {
console.log('calling');
var result = await resolveAfter2Seconds();
console.log(result);
// expected output: 'resolved'
}
asyncCall();
answered 2 hours ago
Negi Rox
1,4591511
Thanks, yes it seems to work! :=) And yeah I know the way you outlined is better, I usually do it that way as well. But this time I have to operate from an AJAX thenable where a series of further AJAX calls is made. Since I want to avoid a huge pyramide of thenables and at the same time cant use the "await" keyword inside a then(), I want to do it this way. Or can you show me how to use the await keyword inside a then? If so, I can open up another question for this, thanks a lot!
– JSONBUG123
2 hours ago
add a comment |
up vote
0
down vote
If we take "fully asynchronous" to mean that there is nothing put on the Event Loop then no, your function a would not become fully synchronous.
Using async/await does not make asynchronous code synchronous, it is simply syntax that helps the developer reason with Promises more easily, essentially by abstracting away the Promise API into the language.
Since I synchronized the AJAX part of function a(), would this mean that at the level of wrapperFunction(), JS does NOT procede until function a() and all of its contents have finished execution?
With the caveat that your function a is not "synchronized" by using async/await, callers of a can still wait for its execution to complete before continuing to the next statement, but only if the call to a is made using the await keyword, like so:
async function a () {
return someAsyncOperation()
}
async function waitsOnA () {
console.log('a started')
await a()
console.log('a completed')
}
// equivalent to
function waitsOnA () {
console.log('a started')
a().then(() => {
console.log('a completed')
})
}
answered 2 hours ago
sdgluck
9,59612546
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The async function declaration defines an asynchronous function, which returns an AsyncFunction object. An asynchronous function is a function which operates asynchronously via the event loop, using an implicit Promise to return its result. But the syntax and structure of your code using async functions is much more like using standard synchronous functions.
So as per your question.
async function a() {
await $.post('someFile.php', {
//some code
})
}
yes it makes it asynchronous. but it is always better to use await where we are calling this function something like below.
for more refrenece please take a look here.
function resolveAfter2Seconds() {
return new Promise(resolve => {
setTimeout(() => {
resolve('resolved');
}, 2000);
});
}
async function asyncCall() {
console.log('calling');
var result = await resolveAfter2Seconds();
console.log(result);
// expected output: 'resolved'
}
asyncCall();
answered 2 hours ago
Negi Rox
1,4591511
Thanks, yes it seems to work! :=) And yeah I know the way you outlined is better, I usually do it that way as well. But this time I have to operate from an AJAX thenable where a series of further AJAX calls is made. Since I want to avoid a huge pyramide of thenables and at the same time cant use the "await" keyword inside a then(), I want to do it this way. Or can you show me how to use the await keyword inside a then? If so, I can open up another question for this, thanks a lot!
– JSONBUG123
2 hours ago
add a comment |
up vote
1
down vote
accepted
The async function declaration defines an asynchronous function, which returns an AsyncFunction object. An asynchronous function is a function which operates asynchronously via the event loop, using an implicit Promise to return its result. But the syntax and structure of your code using async functions is much more like using standard synchronous functions.
So as per your question.
async function a() {
await $.post('someFile.php', {
//some code
})
}
yes it makes it asynchronous. but it is always better to use await where we are calling this function something like below.
for more refrenece please take a look here.
function resolveAfter2Seconds() {
return new Promise(resolve => {
setTimeout(() => {
resolve('resolved');
}, 2000);
});
}
async function asyncCall() {
console.log('calling');
var result = await resolveAfter2Seconds();
console.log(result);
// expected output: 'resolved'
}
asyncCall();
answered 2 hours ago
Negi Rox
1,4591511
Thanks, yes it seems to work! :=) And yeah I know the way you outlined is better, I usually do it that way as well. But this time I have to operate from an AJAX thenable where a series of further AJAX calls is made. Since I want to avoid a huge pyramide of thenables and at the same time cant use the "await" keyword inside a then(), I want to do it this way. Or can you show me how to use the await keyword inside a then? If so, I can open up another question for this, thanks a lot!
– JSONBUG123
2 hours ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The async function declaration defines an asynchronous function, which returns an AsyncFunction object. An asynchronous function is a function which operates asynchronously via the event loop, using an implicit Promise to return its result. But the syntax and structure of your code using async functions is much more like using standard synchronous functions.
So as per your question.
async function a() {
await $.post('someFile.php', {
//some code
})
}
yes it makes it asynchronous. but it is always better to use await where we are calling this function something like below.
for more refrenece please take a look here.
function resolveAfter2Seconds() {
return new Promise(resolve => {
setTimeout(() => {
resolve('resolved');
}, 2000);
});
}
async function asyncCall() {
console.log('calling');
var result = await resolveAfter2Seconds();
console.log(result);
// expected output: 'resolved'
}
asyncCall();
answered 2 hours ago
Negi Rox
1,4591511
The async function declaration defines an asynchronous function, which returns an AsyncFunction object. An asynchronous function is a function which operates asynchronously via the event loop, using an implicit Promise to return its result. But the syntax and structure of your code using async functions is much more like using standard synchronous functions.
So as per your question.
async function a() {
await $.post('someFile.php', {
//some code
})
}
yes it makes it asynchronous. but it is always better to use await where we are calling this function something like below.
for more refrenece please take a look here.
function resolveAfter2Seconds() {
return new Promise(resolve => {
setTimeout(() => {
resolve('resolved');
}, 2000);
});
}
async function asyncCall() {
console.log('calling');
var result = await resolveAfter2Seconds();
console.log(result);
// expected output: 'resolved'
}
asyncCall();
answered 2 hours ago
Negi Rox
1,4591511
edited 2 hours ago
answered 2 hours ago
Negi Rox
1,4591511
answered 2 hours ago
Negi Rox
1,4591511
answered 2 hours ago
Negi Rox
1,4591511
1,4591511
Thanks, yes it seems to work! :=) And yeah I know the way you outlined is better, I usually do it that way as well. But this time I have to operate from an AJAX thenable where a series of further AJAX calls is made. Since I want to avoid a huge pyramide of thenables and at the same time cant use the "await" keyword inside a then(), I want to do it this way. Or can you show me how to use the await keyword inside a then? If so, I can open up another question for this, thanks a lot!
– JSONBUG123
2 hours ago
add a comment |
Thanks, yes it seems to work! :=) And yeah I know the way you outlined is better, I usually do it that way as well. But this time I have to operate from an AJAX thenable where a series of further AJAX calls is made. Since I want to avoid a huge pyramide of thenables and at the same time cant use the "await" keyword inside a then(), I want to do it this way. Or can you show me how to use the await keyword inside a then? If so, I can open up another question for this, thanks a lot!
– JSONBUG123
2 hours ago
Thanks, yes it seems to work! :=) And yeah I know the way you outlined is better, I usually do it that way as well. But this time I have to operate from an AJAX thenable where a series of further AJAX calls is made. Since I want to avoid a huge pyramide of thenables and at the same time cant use the "await" keyword inside a then(), I want to do it this way. Or can you show me how to use the await keyword inside a then? If so, I can open up another question for this, thanks a lot!
– JSONBUG123
2 hours ago
Thanks, yes it seems to work! :=) And yeah I know the way you outlined is better, I usually do it that way as well. But this time I have to operate from an AJAX thenable where a series of further AJAX calls is made. Since I want to avoid a huge pyramide of thenables and at the same time cant use the "await" keyword inside a then(), I want to do it this way. Or can you show me how to use the await keyword inside a then? If so, I can open up another question for this, thanks a lot!
– JSONBUG123
2 hours ago
add a comment |
up vote
0
down vote
If we take "fully asynchronous" to mean that there is nothing put on the Event Loop then no, your function a would not become fully synchronous.
Using async/await does not make asynchronous code synchronous, it is simply syntax that helps the developer reason with Promises more easily, essentially by abstracting away the Promise API into the language.
Since I synchronized the AJAX part of function a(), would this mean that at the level of wrapperFunction(), JS does NOT procede until function a() and all of its contents have finished execution?
With the caveat that your function a is not "synchronized" by using async/await, callers of a can still wait for its execution to complete before continuing to the next statement, but only if the call to a is made using the await keyword, like so:
async function a () {
return someAsyncOperation()
}
async function waitsOnA () {
console.log('a started')
await a()
console.log('a completed')
}
// equivalent to
function waitsOnA () {
console.log('a started')
a().then(() => {
console.log('a completed')
})
}
answered 2 hours ago
sdgluck
9,59612546
add a comment |
up vote
0
down vote
If we take "fully asynchronous" to mean that there is nothing put on the Event Loop then no, your function a would not become fully synchronous.
Using async/await does not make asynchronous code synchronous, it is simply syntax that helps the developer reason with Promises more easily, essentially by abstracting away the Promise API into the language.
Since I synchronized the AJAX part of function a(), would this mean that at the level of wrapperFunction(), JS does NOT procede until function a() and all of its contents have finished execution?
With the caveat that your function a is not "synchronized" by using async/await, callers of a can still wait for its execution to complete before continuing to the next statement, but only if the call to a is made using the await keyword, like so:
async function a () {
return someAsyncOperation()
}
async function waitsOnA () {
console.log('a started')
await a()
console.log('a completed')
}
// equivalent to
function waitsOnA () {
console.log('a started')
a().then(() => {
console.log('a completed')
})
}
answered 2 hours ago
sdgluck
9,59612546
add a comment |
up vote
0
down vote
up vote
0
down vote
If we take "fully asynchronous" to mean that there is nothing put on the Event Loop then no, your function a would not become fully synchronous.
Using async/await does not make asynchronous code synchronous, it is simply syntax that helps the developer reason with Promises more easily, essentially by abstracting away the Promise API into the language.
Since I synchronized the AJAX part of function a(), would this mean that at the level of wrapperFunction(), JS does NOT procede until function a() and all of its contents have finished execution?
With the caveat that your function a is not "synchronized" by using async/await, callers of a can still wait for its execution to complete before continuing to the next statement, but only if the call to a is made using the await keyword, like so:
async function a () {
return someAsyncOperation()
}
async function waitsOnA () {
console.log('a started')
await a()
console.log('a completed')
}
// equivalent to
function waitsOnA () {
console.log('a started')
a().then(() => {
console.log('a completed')
})
}
answered 2 hours ago
sdgluck
9,59612546
If we take "fully asynchronous" to mean that there is nothing put on the Event Loop then no, your function a would not become fully synchronous.
Using async/await does not make asynchronous code synchronous, it is simply syntax that helps the developer reason with Promises more easily, essentially by abstracting away the Promise API into the language.
Since I synchronized the AJAX part of function a(), would this mean that at the level of wrapperFunction(), JS does NOT procede until function a() and all of its contents have finished execution?
With the caveat that your function a is not "synchronized" by using async/await, callers of a can still wait for its execution to complete before continuing to the next statement, but only if the call to a is made using the await keyword, like so:
async function a () {
return someAsyncOperation()
}
async function waitsOnA () {
console.log('a started')
await a()
console.log('a completed')
}
// equivalent to
function waitsOnA () {
console.log('a started')
a().then(() => {
console.log('a completed')
})
}
answered 2 hours ago
sdgluck
9,59612546
answered 2 hours ago
sdgluck
9,59612546
answered 2 hours ago
sdgluck
9,59612546
answered 2 hours ago
sdgluck
9,59612546
9,59612546
add a comment |
add a comment |
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1
It might help to look at the
awaitcode and think of what it would look like when it was written with.then()– Bergi
3 hours ago
You need to write
await a()inwrapperFunction()to make it seem like it's synchronous. And then you need to declareasync function wrapperFunction()– Barmar
3 hours ago
@Barmar I know this way, but out of curiosity, I wanted to know whether the way I described would work as well or not.
– JSONBUG123
3 hours ago
The simple answer is no. Declaring a function
asyncis just syntactic sugar for automatically returning a promise. It doesn't make it act synchronous. Theawaitkeyword is what makes a call to an async function seem like it's synchronous.– Barmar
3 hours ago
but the keyword "async" doesn't make a function return promises? Instead, an async function enables the await keyword for use INSIDE the async function, which can await promises being returned from any function inside the function which was prefixed with "async" .
– JSONBUG123
3 hours ago