Pink Hat
Pink Hat17mo ago

✅ – Pink Hat – 00-54 Feb 15

in this code
this.pattern = [
{
method: new Function(`console.log(1)`),
bulletDelay: 1000
} ,{
method: new Function(`console.log(2)`),
bulletDelay: 1000

} ,{
method: new Function(`console.log(3)`),
bulletDelay: 1000

}
]

for (this.i in this.pattern) {

setTimeout(()=>this.pattern[this.i].method(), function (i){
this.z = 0;
for (this.j = i-1; this.j >=0; this.j--) {
try {
this.z += this.pattern[this.j]?.bulletDelay
} catch (error) {
console.error(error)
} finally {
continue
}
}
console.log(i)
return this.z

}(this.i))
console.log(this.pattern[this.i])

}
this.pattern = [
{
method: new Function(`console.log(1)`),
bulletDelay: 1000
} ,{
method: new Function(`console.log(2)`),
bulletDelay: 1000

} ,{
method: new Function(`console.log(3)`),
bulletDelay: 1000

}
]

for (this.i in this.pattern) {

setTimeout(()=>this.pattern[this.i].method(), function (i){
this.z = 0;
for (this.j = i-1; this.j >=0; this.j--) {
try {
this.z += this.pattern[this.j]?.bulletDelay
} catch (error) {
console.error(error)
} finally {
continue
}
}
console.log(i)
return this.z

}(this.i))
console.log(this.pattern[this.i])

}
why does this.pattern[2] get printed 3 times?
> 0
> {bulletDelay: 1000, method: ƒ}
> 1
> {bulletDelay: 1000, method: ƒ}
> 2
> {bulletDelay: 1000, method: ƒ}
> undefined
> 3
> 3
> 3
> 0
> {bulletDelay: 1000, method: ƒ}
> 1
> {bulletDelay: 1000, method: ƒ}
> 2
> {bulletDelay: 1000, method: ƒ}
> undefined
> 3
> 3
> 3
When it should be printing
> 0
> {bulletDelay: 1000, method: ƒ}
> 1
> {bulletDelay: 1000, method: ƒ}
> 2
> {bulletDelay: 1000, method: ƒ}
> undefined
> 1
> 2
> 3
> 0
> {bulletDelay: 1000, method: ƒ}
> 1
> {bulletDelay: 1000, method: ƒ}
> 2
> {bulletDelay: 1000, method: ƒ}
> undefined
> 1
> 2
> 3
46 Replies
ScriptyChris
ScriptyChris17mo ago
Try caching this.j as a local variable
Pink Hat
Pink Hat17mo ago
caching it?
ScriptyChris
ScriptyChris17mo ago
Yes, similarly as you do with i regarding this.i
Pink Hat
Pink Hat17mo ago
like this?
for (this.i in this.pattern) {

setTimeout(()=>this.pattern[this.i].method(), function (i, j){
this.z = 0;
for (j >=0; j--) {
try {
this.z += this.pattern[j]?.bulletDelay
} catch (error) {
console.error(error)
} finally {
continue
}
}
console.log(i)
return this.z

}(this.i, 0))
console.log(this.pattern[this.i])

}
for (this.i in this.pattern) {

setTimeout(()=>this.pattern[this.i].method(), function (i, j){
this.z = 0;
for (j >=0; j--) {
try {
this.z += this.pattern[j]?.bulletDelay
} catch (error) {
console.error(error)
} finally {
continue
}
}
console.log(i)
return this.z

}(this.i, 0))
console.log(this.pattern[this.i])

}
ScriptyChris
ScriptyChris17mo ago
}(this.i, 0)) -> }(this.i, this.j))
Pink Hat
Pink Hat17mo ago
nothing still the same result
ScriptyChris
ScriptyChris17mo ago
Can you add labels to console.log so it's more readable, which log refers to what value?
Pink Hat
Pink Hat17mo ago
sure
ScriptyChris
ScriptyChris17mo ago
console.log('i:', i)

console.log('this.pattern[this.i]:',this.pattern[this.i])
console.log('i:', i)

console.log('this.pattern[this.i]:',this.pattern[this.i])
And show output
Pink Hat
Pink Hat17mo ago
ScriptyChris
ScriptyChris17mo ago
Where's that "3" log coming from? Add it label too
Pink Hat
Pink Hat17mo ago
the 3 is the label it was originally this
Pink Hat
Pink Hat17mo ago
Pink Hat
Pink Hat17mo ago
then so I could run it in chrome
this.pattern = [
{
method: new Function(`console.log(1)`),
bulletDelay: 1000
} ,{
method: new Function(`console.log(2)`),
bulletDelay: 1000

} ,{
method: new Function(`console.log(3)`),
bulletDelay: 1000

}
]
this.pattern = [
{
method: new Function(`console.log(1)`),
bulletDelay: 1000
} ,{
method: new Function(`console.log(2)`),
bulletDelay: 1000

} ,{
method: new Function(`console.log(3)`),
bulletDelay: 1000

}
]
Its a result of this.pattern[2] being run 3 times I think it's happening because this.i is already fully changed before the setTimeout occurs
ScriptyChris
ScriptyChris17mo ago
Can you add counters to pattern too?
this.pattern = [
{
method: new Function(`console.log(1)`),
bulletDelay: 1000 ,
counter: 1,
} ,{
method: new Function(`console.log(2)`),
bulletDelay: 1000,
counter: 2,
} ,{
method: new Function(`console.log(3)`),
bulletDelay: 1000,
counter: 3,
}
]
this.pattern = [
{
method: new Function(`console.log(1)`),
bulletDelay: 1000 ,
counter: 1,
} ,{
method: new Function(`console.log(2)`),
bulletDelay: 1000,
counter: 2,
} ,{
method: new Function(`console.log(3)`),
bulletDelay: 1000,
counter: 3,
}
]
Show the console output - i suppose each pattern will have counter: 3 Yes You could await timeout call
Pink Hat
Pink Hat17mo ago
in the for function?
ScriptyChris
ScriptyChris17mo ago
for (this.i in this.pattern) {

await new Promise(resolve => setTimeout(()=>this.pattern[this.i].method(), function (i, j){
// stuff

// `return this.z` becomes
resolve(this.z)
}(this.i, 0))
console.log(this.pattern[this.i])
}
for (this.i in this.pattern) {

await new Promise(resolve => setTimeout(()=>this.pattern[this.i].method(), function (i, j){
// stuff

// `return this.z` becomes
resolve(this.z)
}(this.i, 0))
console.log(this.pattern[this.i])
}
Pink Hat
Pink Hat17mo ago
would that work?
ScriptyChris
ScriptyChris17mo ago
It should
Pink Hat
Pink Hat17mo ago
ok. I'll try it
ScriptyChris
ScriptyChris17mo ago
If this for..in is inside a function, then you have to make it async If it's in top-level scope, then your script should work as a module, otherwise await won't work, and you have to .then() the promise
Pink Hat
Pink Hat17mo ago
and inside the .then you would put what exactly? because it seems the await doesnt have anything referencing it
ScriptyChris
ScriptyChris17mo ago
Your console.log
Pink Hat
Pink Hat17mo ago
ok.
ScriptyChris
ScriptyChris17mo ago
What do you mean?
Pink Hat
Pink Hat17mo ago
nvm that so
Pink Hat
Pink Hat17mo ago
Pink Hat
Pink Hat17mo ago
now it's worse it doesnt even wait the time it needs to
ScriptyChris
ScriptyChris17mo ago
No, you have to wrap console.log inside a function, which you pass to .then() https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then#using_the_then_method .then(), same as setTimeout() or Promise() expects a function to be passed, which is then called inside
Pink Hat
Pink Hat17mo ago
huh? what do u mean by this im confused as hell
ScriptyChris
ScriptyChris17mo ago
Read example in the docs
ScriptyChris
ScriptyChris17mo ago
Passing a function
ScriptyChris
ScriptyChris17mo ago
.then((value) => console.log( /* stuff */ ))
Pink Hat
Pink Hat17mo ago
I want the promise to pass after a certain time. if I do
new Promise(resolve).then(() => console.log(/* stuff */)
new Promise(resolve).then(() => console.log(/* stuff */)
then what does that do
ScriptyChris
ScriptyChris17mo ago
What resolve is there? I already wrote you example with promisified setTimeout above ..ok, try this - i think i misread your code (2:30am on my side)
await new Promise(
(resolve) => {
setTimeout(
() => {
this.pattern[this.i].method();
resolve();
},
function (i, j){
// stuff

return this.z;
}(this.i, 0)
)
}
)
console.log(this.pattern[this.i])
await new Promise(
(resolve) => {
setTimeout(
() => {
this.pattern[this.i].method();
resolve();
},
function (i, j){
// stuff

return this.z;
}(this.i, 0)
)
}
)
console.log(this.pattern[this.i])
Pink Hat
Pink Hat17mo ago
threw this error (after I fixed some of the syntax cus it was complaining
ScriptyChris
ScriptyChris17mo ago
Lacking closing curly brace
ScriptyChris
ScriptyChris17mo ago
Add } there
ScriptyChris
ScriptyChris17mo ago
And the console.log('what') is incorrectly place there Just copy-paste this ☝️
ScriptyChris
ScriptyChris17mo ago
Works for me on dry-run
Pink Hat
Pink Hat17mo ago
it works
ScriptyChris
ScriptyChris17mo ago
👍
Pink Hat
Pink Hat17mo ago
I found a piece of code that works (slightly waltered from yours)
this.pattern = [
{
method: new Function('console.log("Counter: 1")'),
bulletDelay: 1000
} ,{
method: new Function('console.log("Counter: 2")'),
bulletDelay: 1000

} ,{
method: new Function('console.log("Counter: 3")'),
bulletDelay: 1000

}
]

for (this.i in this.pattern) {
await new Promise(
(resolve) => {
setTimeout(
() => {
this.pattern[this.i].method();
resolve();
},
function (i, j){
// stuff

return this.z;
}(this.i, 0)
)
console.log(this.pattern[this.i])
});
}
this.pattern = [
{
method: new Function('console.log("Counter: 1")'),
bulletDelay: 1000
} ,{
method: new Function('console.log("Counter: 2")'),
bulletDelay: 1000

} ,{
method: new Function('console.log("Counter: 3")'),
bulletDelay: 1000

}
]

for (this.i in this.pattern) {
await new Promise(
(resolve) => {
setTimeout(
() => {
this.pattern[this.i].method();
resolve();
},
function (i, j){
// stuff

return this.z;
}(this.i, 0)
)
console.log(this.pattern[this.i])
});
}
and now we can close this. thanks for the help @ScriptyChris #thanks
ScriptyChris
ScriptyChris17mo ago
You may react to helpful answer with ✅ to mark thread as solved
reactibot
reactibot17mo ago
This question has an answer! Thank you for helping 😄 If you have a followup question, you may want to reply to this thread so other members know they're related. https://discord.com/channels/102860784329052160/565213527673929729/1075218422025158786
reactibot
reactibot17mo ago
This thread hasn’t had any activity in 12 hours, so it’s now locked. Threads are closed automatically after 12 hours. If you have a followup question, you may want to reply to this thread so other members know they're related. https://discord.com/channels/102860784329052160/565213527673929729/1075218422025158786 Question not getting answered? Maybe it's hard to answer, or maybe you asked at a slow time. Check out these resources for help asking a good question: https://stackoverflow.com/help/how-to-ask http://wp.me/p2oIwo-26