Retry promise until resolved (Too much recursion error)
I was trying to test if I could have a promise retry until it resolved, but it keeps giving "too much recursion" error and I can't understand why it doesn't stop after the 3rd recursion.
The following code is trying to emulate a failed query request to a server, which comes from a non-promise function.
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) => {
nonPromiseCallback(x, resolve, tryUntilThree(x+1));
})
}
tryUntilThree(1)
.then(console.log);
javascript ecmascript-6 es6-promise
add a comment |
I was trying to test if I could have a promise retry until it resolved, but it keeps giving "too much recursion" error and I can't understand why it doesn't stop after the 3rd recursion.
The following code is trying to emulate a failed query request to a server, which comes from a non-promise function.
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) => {
nonPromiseCallback(x, resolve, tryUntilThree(x+1));
})
}
tryUntilThree(1)
.then(console.log);
javascript ecmascript-6 es6-promise
2
check this answer, might help you
– The Reason
Jan 18 at 13:28
You get too much recursion because tryUntilThree is called too many times. Notice that you have written tryUntilThree(x+1), ie the engine has to resolve the call to tryUntilThree before it can call nonPromiseCallback. You have an infinite loop there.
– some
Jan 18 at 13:33
@some: Mind if i quote your comment in my answer? :)
– vicbyte
Jan 18 at 13:48
@vicbyte I don't mind. I added special quotes around the code (in my comment they where in italics)
– some
Jan 19 at 1:20
add a comment |
I was trying to test if I could have a promise retry until it resolved, but it keeps giving "too much recursion" error and I can't understand why it doesn't stop after the 3rd recursion.
The following code is trying to emulate a failed query request to a server, which comes from a non-promise function.
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) => {
nonPromiseCallback(x, resolve, tryUntilThree(x+1));
})
}
tryUntilThree(1)
.then(console.log);
javascript ecmascript-6 es6-promise
I was trying to test if I could have a promise retry until it resolved, but it keeps giving "too much recursion" error and I can't understand why it doesn't stop after the 3rd recursion.
The following code is trying to emulate a failed query request to a server, which comes from a non-promise function.
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) => {
nonPromiseCallback(x, resolve, tryUntilThree(x+1));
})
}
tryUntilThree(1)
.then(console.log);
javascript ecmascript-6 es6-promise
javascript ecmascript-6 es6-promise
edited Jan 18 at 13:26
Mojimi
asked Jan 18 at 13:02
MojimiMojimi
247831
247831
2
check this answer, might help you
– The Reason
Jan 18 at 13:28
You get too much recursion because tryUntilThree is called too many times. Notice that you have written tryUntilThree(x+1), ie the engine has to resolve the call to tryUntilThree before it can call nonPromiseCallback. You have an infinite loop there.
– some
Jan 18 at 13:33
@some: Mind if i quote your comment in my answer? :)
– vicbyte
Jan 18 at 13:48
@vicbyte I don't mind. I added special quotes around the code (in my comment they where in italics)
– some
Jan 19 at 1:20
add a comment |
2
check this answer, might help you
– The Reason
Jan 18 at 13:28
You get too much recursion because tryUntilThree is called too many times. Notice that you have written tryUntilThree(x+1), ie the engine has to resolve the call to tryUntilThree before it can call nonPromiseCallback. You have an infinite loop there.
– some
Jan 18 at 13:33
@some: Mind if i quote your comment in my answer? :)
– vicbyte
Jan 18 at 13:48
@vicbyte I don't mind. I added special quotes around the code (in my comment they where in italics)
– some
Jan 19 at 1:20
2
2
check this answer, might help you
– The Reason
Jan 18 at 13:28
check this answer, might help you
– The Reason
Jan 18 at 13:28
You get too much recursion because tryUntilThree is called too many times. Notice that you have written tryUntilThree(x+1), ie the engine has to resolve the call to tryUntilThree before it can call nonPromiseCallback. You have an infinite loop there.
– some
Jan 18 at 13:33
You get too much recursion because tryUntilThree is called too many times. Notice that you have written tryUntilThree(x+1), ie the engine has to resolve the call to tryUntilThree before it can call nonPromiseCallback. You have an infinite loop there.
– some
Jan 18 at 13:33
@some: Mind if i quote your comment in my answer? :)
– vicbyte
Jan 18 at 13:48
@some: Mind if i quote your comment in my answer? :)
– vicbyte
Jan 18 at 13:48
@vicbyte I don't mind. I added special quotes around the code (in my comment they where in italics)
– some
Jan 19 at 1:20
@vicbyte I don't mind. I added special quotes around the code (in my comment they where in italics)
– some
Jan 19 at 1:20
add a comment |
4 Answers
4
active
oldest
votes
Since you are interested in a promise failure, you could use the catch handler.
As to why your code doesn't work, heres a good explaination by some (also in the comment):
You get too much recursion because
tryUntilThreeis called too many
times. Notice that you have writtentryUntilThree(x+1), ie the engine
has to resolve the call totryUntilThreebefore it can call
nonPromiseCallback. You have an infinite loop there.
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) =>
nonPromiseCallback(x, resolve, reject)
).catch(() =>
tryUntilThree(x + 1)
)
}
tryUntilThree(1)
.then(console.log);add a comment |
The problem is with the nonPromiseCallback method call, instead of a function reference, you are passing an actual function and then calling that function.
Problem:
nonPromiseCallback(x, resolve, tryUntilThree(x+1));
Fix:
nonPromiseCallback(x, resolve, tryUntilThree);
and
reject(x+1);
This doesn't actually work either.
– Bergi
Jan 18 at 13:37
@Bergi I pointed the fix for the call stack issue he is facing (infinite recursion), I haven't pointed out any thing else since I don't know what he wants from the code.
– Danyal Imran
Jan 18 at 13:38
I want the code to compile... and to print 3
– Mojimi
Jan 18 at 13:43
add a comment |
I tried your code but got the TypeError: reject is not a function.
That is because you pass in tryUntilThree(x+1) which executes the function before passing it to nonPromiseCallback.
So i came up with this code, to try to accomplish what you want.
let res; // used for keeping a reference to the original resolve
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x + 1);
}
else{
resolve(x);
}
}
function tryUntilThree(x){
return new Promise((resolve) => {
if(!res){
res = resolve;
}
nonPromiseCallback(x, res, tryUntilThree);
});
}
tryUntilThree(1)
.then(res => console.log("RESULT:", res));
let res;, this variable is used for keeping a reference to the original resolve, so that the .then executes.
add a comment |
As I said in my comment, to your question, you had an infinite loop since to call nonPromiseCallback the result of tryUntilThree was needed... and to get that, tryUntilThree was called... and it was looping until memory was exhausted or the end of the universe, whichever came first.
You need to make two changes:
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1) // the increment of x is moved to here.
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) => {
nonPromiseCallback(x, resolve, tryUntilThree); // Just pass the function, it will be called later
})
}
tryUntilThree(1)
.then(console.log);
If you can use async and await (new features since 2017) you could solve it like this (I moved the decision of maximum number of tries to the calling function) :
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1)
}else{
resolve(x)
}
}
async function tryUntilThree(){
const maxAttempts = 3;
let attempt = 1;
do {
try {
// "await" waits for the prommise to resolve or reject.
// If it is rejected, an error will be thrown. That's why
// this part of the code is inside a try/catch-block.
const result = await new Promise( (resolve, reject) =>
nonPromiseCallback( attempt, resolve, reject )
);
return result; // Return the result from nonPromiseCallback
}
catch (error) {
// The nonPromiseCallback failed. Try again
attempt += 1;
}
} while ( attempt <= maxAttempts );
// Signal error after all retires.
throw Error(`Failure, even after ${maxAttempts} tries`);
}
tryUntilThree()
.then(console.log);
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since you are interested in a promise failure, you could use the catch handler.
As to why your code doesn't work, heres a good explaination by some (also in the comment):
You get too much recursion because
tryUntilThreeis called too many
times. Notice that you have writtentryUntilThree(x+1), ie the engine
has to resolve the call totryUntilThreebefore it can call
nonPromiseCallback. You have an infinite loop there.
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) =>
nonPromiseCallback(x, resolve, reject)
).catch(() =>
tryUntilThree(x + 1)
)
}
tryUntilThree(1)
.then(console.log);add a comment |
Since you are interested in a promise failure, you could use the catch handler.
As to why your code doesn't work, heres a good explaination by some (also in the comment):
You get too much recursion because
tryUntilThreeis called too many
times. Notice that you have writtentryUntilThree(x+1), ie the engine
has to resolve the call totryUntilThreebefore it can call
nonPromiseCallback. You have an infinite loop there.
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) =>
nonPromiseCallback(x, resolve, reject)
).catch(() =>
tryUntilThree(x + 1)
)
}
tryUntilThree(1)
.then(console.log);add a comment |
Since you are interested in a promise failure, you could use the catch handler.
As to why your code doesn't work, heres a good explaination by some (also in the comment):
You get too much recursion because
tryUntilThreeis called too many
times. Notice that you have writtentryUntilThree(x+1), ie the engine
has to resolve the call totryUntilThreebefore it can call
nonPromiseCallback. You have an infinite loop there.
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) =>
nonPromiseCallback(x, resolve, reject)
).catch(() =>
tryUntilThree(x + 1)
)
}
tryUntilThree(1)
.then(console.log);Since you are interested in a promise failure, you could use the catch handler.
As to why your code doesn't work, heres a good explaination by some (also in the comment):
You get too much recursion because
tryUntilThreeis called too many
times. Notice that you have writtentryUntilThree(x+1), ie the engine
has to resolve the call totryUntilThreebefore it can call
nonPromiseCallback. You have an infinite loop there.
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) =>
nonPromiseCallback(x, resolve, reject)
).catch(() =>
tryUntilThree(x + 1)
)
}
tryUntilThree(1)
.then(console.log);function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) =>
nonPromiseCallback(x, resolve, reject)
).catch(() =>
tryUntilThree(x + 1)
)
}
tryUntilThree(1)
.then(console.log);function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x)
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) =>
nonPromiseCallback(x, resolve, reject)
).catch(() =>
tryUntilThree(x + 1)
)
}
tryUntilThree(1)
.then(console.log);edited Jan 19 at 1:19
some
35.4k116479
35.4k116479
answered Jan 18 at 13:44
vicbytevicbyte
2,6011313
2,6011313
add a comment |
add a comment |
The problem is with the nonPromiseCallback method call, instead of a function reference, you are passing an actual function and then calling that function.
Problem:
nonPromiseCallback(x, resolve, tryUntilThree(x+1));
Fix:
nonPromiseCallback(x, resolve, tryUntilThree);
and
reject(x+1);
This doesn't actually work either.
– Bergi
Jan 18 at 13:37
@Bergi I pointed the fix for the call stack issue he is facing (infinite recursion), I haven't pointed out any thing else since I don't know what he wants from the code.
– Danyal Imran
Jan 18 at 13:38
I want the code to compile... and to print 3
– Mojimi
Jan 18 at 13:43
add a comment |
The problem is with the nonPromiseCallback method call, instead of a function reference, you are passing an actual function and then calling that function.
Problem:
nonPromiseCallback(x, resolve, tryUntilThree(x+1));
Fix:
nonPromiseCallback(x, resolve, tryUntilThree);
and
reject(x+1);
This doesn't actually work either.
– Bergi
Jan 18 at 13:37
@Bergi I pointed the fix for the call stack issue he is facing (infinite recursion), I haven't pointed out any thing else since I don't know what he wants from the code.
– Danyal Imran
Jan 18 at 13:38
I want the code to compile... and to print 3
– Mojimi
Jan 18 at 13:43
add a comment |
The problem is with the nonPromiseCallback method call, instead of a function reference, you are passing an actual function and then calling that function.
Problem:
nonPromiseCallback(x, resolve, tryUntilThree(x+1));
Fix:
nonPromiseCallback(x, resolve, tryUntilThree);
and
reject(x+1);
The problem is with the nonPromiseCallback method call, instead of a function reference, you are passing an actual function and then calling that function.
Problem:
nonPromiseCallback(x, resolve, tryUntilThree(x+1));
Fix:
nonPromiseCallback(x, resolve, tryUntilThree);
and
reject(x+1);
answered Jan 18 at 13:36
Danyal ImranDanyal Imran
1,145214
1,145214
This doesn't actually work either.
– Bergi
Jan 18 at 13:37
@Bergi I pointed the fix for the call stack issue he is facing (infinite recursion), I haven't pointed out any thing else since I don't know what he wants from the code.
– Danyal Imran
Jan 18 at 13:38
I want the code to compile... and to print 3
– Mojimi
Jan 18 at 13:43
add a comment |
This doesn't actually work either.
– Bergi
Jan 18 at 13:37
@Bergi I pointed the fix for the call stack issue he is facing (infinite recursion), I haven't pointed out any thing else since I don't know what he wants from the code.
– Danyal Imran
Jan 18 at 13:38
I want the code to compile... and to print 3
– Mojimi
Jan 18 at 13:43
This doesn't actually work either.
– Bergi
Jan 18 at 13:37
This doesn't actually work either.
– Bergi
Jan 18 at 13:37
@Bergi I pointed the fix for the call stack issue he is facing (infinite recursion), I haven't pointed out any thing else since I don't know what he wants from the code.
– Danyal Imran
Jan 18 at 13:38
@Bergi I pointed the fix for the call stack issue he is facing (infinite recursion), I haven't pointed out any thing else since I don't know what he wants from the code.
– Danyal Imran
Jan 18 at 13:38
I want the code to compile... and to print 3
– Mojimi
Jan 18 at 13:43
I want the code to compile... and to print 3
– Mojimi
Jan 18 at 13:43
add a comment |
I tried your code but got the TypeError: reject is not a function.
That is because you pass in tryUntilThree(x+1) which executes the function before passing it to nonPromiseCallback.
So i came up with this code, to try to accomplish what you want.
let res; // used for keeping a reference to the original resolve
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x + 1);
}
else{
resolve(x);
}
}
function tryUntilThree(x){
return new Promise((resolve) => {
if(!res){
res = resolve;
}
nonPromiseCallback(x, res, tryUntilThree);
});
}
tryUntilThree(1)
.then(res => console.log("RESULT:", res));
let res;, this variable is used for keeping a reference to the original resolve, so that the .then executes.
add a comment |
I tried your code but got the TypeError: reject is not a function.
That is because you pass in tryUntilThree(x+1) which executes the function before passing it to nonPromiseCallback.
So i came up with this code, to try to accomplish what you want.
let res; // used for keeping a reference to the original resolve
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x + 1);
}
else{
resolve(x);
}
}
function tryUntilThree(x){
return new Promise((resolve) => {
if(!res){
res = resolve;
}
nonPromiseCallback(x, res, tryUntilThree);
});
}
tryUntilThree(1)
.then(res => console.log("RESULT:", res));
let res;, this variable is used for keeping a reference to the original resolve, so that the .then executes.
add a comment |
I tried your code but got the TypeError: reject is not a function.
That is because you pass in tryUntilThree(x+1) which executes the function before passing it to nonPromiseCallback.
So i came up with this code, to try to accomplish what you want.
let res; // used for keeping a reference to the original resolve
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x + 1);
}
else{
resolve(x);
}
}
function tryUntilThree(x){
return new Promise((resolve) => {
if(!res){
res = resolve;
}
nonPromiseCallback(x, res, tryUntilThree);
});
}
tryUntilThree(1)
.then(res => console.log("RESULT:", res));
let res;, this variable is used for keeping a reference to the original resolve, so that the .then executes.
I tried your code but got the TypeError: reject is not a function.
That is because you pass in tryUntilThree(x+1) which executes the function before passing it to nonPromiseCallback.
So i came up with this code, to try to accomplish what you want.
let res; // used for keeping a reference to the original resolve
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x + 1);
}
else{
resolve(x);
}
}
function tryUntilThree(x){
return new Promise((resolve) => {
if(!res){
res = resolve;
}
nonPromiseCallback(x, res, tryUntilThree);
});
}
tryUntilThree(1)
.then(res => console.log("RESULT:", res));
let res;, this variable is used for keeping a reference to the original resolve, so that the .then executes.
answered Jan 18 at 13:41
robbannnrobbannn
3,41111935
3,41111935
add a comment |
add a comment |
As I said in my comment, to your question, you had an infinite loop since to call nonPromiseCallback the result of tryUntilThree was needed... and to get that, tryUntilThree was called... and it was looping until memory was exhausted or the end of the universe, whichever came first.
You need to make two changes:
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1) // the increment of x is moved to here.
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) => {
nonPromiseCallback(x, resolve, tryUntilThree); // Just pass the function, it will be called later
})
}
tryUntilThree(1)
.then(console.log);
If you can use async and await (new features since 2017) you could solve it like this (I moved the decision of maximum number of tries to the calling function) :
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1)
}else{
resolve(x)
}
}
async function tryUntilThree(){
const maxAttempts = 3;
let attempt = 1;
do {
try {
// "await" waits for the prommise to resolve or reject.
// If it is rejected, an error will be thrown. That's why
// this part of the code is inside a try/catch-block.
const result = await new Promise( (resolve, reject) =>
nonPromiseCallback( attempt, resolve, reject )
);
return result; // Return the result from nonPromiseCallback
}
catch (error) {
// The nonPromiseCallback failed. Try again
attempt += 1;
}
} while ( attempt <= maxAttempts );
// Signal error after all retires.
throw Error(`Failure, even after ${maxAttempts} tries`);
}
tryUntilThree()
.then(console.log);
add a comment |
As I said in my comment, to your question, you had an infinite loop since to call nonPromiseCallback the result of tryUntilThree was needed... and to get that, tryUntilThree was called... and it was looping until memory was exhausted or the end of the universe, whichever came first.
You need to make two changes:
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1) // the increment of x is moved to here.
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) => {
nonPromiseCallback(x, resolve, tryUntilThree); // Just pass the function, it will be called later
})
}
tryUntilThree(1)
.then(console.log);
If you can use async and await (new features since 2017) you could solve it like this (I moved the decision of maximum number of tries to the calling function) :
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1)
}else{
resolve(x)
}
}
async function tryUntilThree(){
const maxAttempts = 3;
let attempt = 1;
do {
try {
// "await" waits for the prommise to resolve or reject.
// If it is rejected, an error will be thrown. That's why
// this part of the code is inside a try/catch-block.
const result = await new Promise( (resolve, reject) =>
nonPromiseCallback( attempt, resolve, reject )
);
return result; // Return the result from nonPromiseCallback
}
catch (error) {
// The nonPromiseCallback failed. Try again
attempt += 1;
}
} while ( attempt <= maxAttempts );
// Signal error after all retires.
throw Error(`Failure, even after ${maxAttempts} tries`);
}
tryUntilThree()
.then(console.log);
add a comment |
As I said in my comment, to your question, you had an infinite loop since to call nonPromiseCallback the result of tryUntilThree was needed... and to get that, tryUntilThree was called... and it was looping until memory was exhausted or the end of the universe, whichever came first.
You need to make two changes:
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1) // the increment of x is moved to here.
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) => {
nonPromiseCallback(x, resolve, tryUntilThree); // Just pass the function, it will be called later
})
}
tryUntilThree(1)
.then(console.log);
If you can use async and await (new features since 2017) you could solve it like this (I moved the decision of maximum number of tries to the calling function) :
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1)
}else{
resolve(x)
}
}
async function tryUntilThree(){
const maxAttempts = 3;
let attempt = 1;
do {
try {
// "await" waits for the prommise to resolve or reject.
// If it is rejected, an error will be thrown. That's why
// this part of the code is inside a try/catch-block.
const result = await new Promise( (resolve, reject) =>
nonPromiseCallback( attempt, resolve, reject )
);
return result; // Return the result from nonPromiseCallback
}
catch (error) {
// The nonPromiseCallback failed. Try again
attempt += 1;
}
} while ( attempt <= maxAttempts );
// Signal error after all retires.
throw Error(`Failure, even after ${maxAttempts} tries`);
}
tryUntilThree()
.then(console.log);
As I said in my comment, to your question, you had an infinite loop since to call nonPromiseCallback the result of tryUntilThree was needed... and to get that, tryUntilThree was called... and it was looping until memory was exhausted or the end of the universe, whichever came first.
You need to make two changes:
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1) // the increment of x is moved to here.
}else{
resolve(x)
}
}
function tryUntilThree(x){
return new Promise( (resolve, reject) => {
nonPromiseCallback(x, resolve, tryUntilThree); // Just pass the function, it will be called later
})
}
tryUntilThree(1)
.then(console.log);
If you can use async and await (new features since 2017) you could solve it like this (I moved the decision of maximum number of tries to the calling function) :
function nonPromiseCallback(x, resolve, reject){
if(x < 3){
reject(x+1)
}else{
resolve(x)
}
}
async function tryUntilThree(){
const maxAttempts = 3;
let attempt = 1;
do {
try {
// "await" waits for the prommise to resolve or reject.
// If it is rejected, an error will be thrown. That's why
// this part of the code is inside a try/catch-block.
const result = await new Promise( (resolve, reject) =>
nonPromiseCallback( attempt, resolve, reject )
);
return result; // Return the result from nonPromiseCallback
}
catch (error) {
// The nonPromiseCallback failed. Try again
attempt += 1;
}
} while ( attempt <= maxAttempts );
// Signal error after all retires.
throw Error(`Failure, even after ${maxAttempts} tries`);
}
tryUntilThree()
.then(console.log);
answered Jan 19 at 2:01
somesome
35.4k116479
35.4k116479
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2
check this answer, might help you
– The Reason
Jan 18 at 13:28
You get too much recursion because tryUntilThree is called too many times. Notice that you have written tryUntilThree(x+1), ie the engine has to resolve the call to tryUntilThree before it can call nonPromiseCallback. You have an infinite loop there.
– some
Jan 18 at 13:33
@some: Mind if i quote your comment in my answer? :)
– vicbyte
Jan 18 at 13:48
@vicbyte I don't mind. I added special quotes around the code (in my comment they where in italics)
– some
Jan 19 at 1:20