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Javascript for loop variable scope [duplicate]

This question already has answers here:
Explanation of `let` and block scoping with for loops
(5 answers)
JavaScript closure inside loops – simple practical example
(44 answers)
let keyword in the for loop
(3 answers)
JavaScript: Understanding let scope inside for loop [duplicate]
(3 answers)
Closed 6 months ago.
This question is inspired by a very famous trick interview question for javascript closures, however I am more interested into another aspect of it. The code goes as follows:
let i
for (i = 0; i < 3; i++) {
setTimeout(function log() {
console.log(i); // What is logged?
}, 1000);
}
The answer is 3 3 3, and (I think) I know the explanation, each callback function can access the same i variable, by the time they are executed its value is 3.
When we write let inside for loop like this:
for (let i = 0; i < 3; i++) {
setTimeout(function log() {
console.log(i); // What is logged?
}, 1000);
}
Now we get 0 1 2. Ok, here the explanation is that when declared this way, i is scoped to this loop, not available outside, and each iteration has its copy of i.
The part that bugs me: it does not seem to be a different variable in each step. If we change i in a first step, it will be changed for other steps.
for (let i = 0; i < 3; i++) {
setTimeout(function log() {
console.log(i); // What is logged?
}, 1000);
if (i === 0) i ++
}
The result now is 1 2, skipped one step.
So I see two possible explanations for this, but I cannot confirm this anywhere.
Closures work differently for block-scoped and function or global scoped variable
After each step of the loop, the current state of the variable declared in a loop is copied to a new variable for the next step of the loop
???
Note: Please let's not focus on why values are 3 3 3 instead of 2 2 2, let's not go into event loop with setTimeout etc. unless relevant to the answer. I feel like closures and variable scopes will be most relevant here.

Does capturing a block-scoped variable cause it to be copied? [duplicate]

This question already has answers here:
JavaScript closure inside loops – simple practical example
(44 answers)
Explanation of `let` and block scoping with for loops
(5 answers)
Closed 2 years ago.
Consider the following code:
const captured = [];
for (var i = 0; i < 2; ++i) {
captured.push(() => console.log(i));
}
for (let x = 0; x < 2; ++x) {
captured.push(() => console.log(x));
}
captured.forEach(v => v());
This gives the above output in the snippet.
In both cases, we effectively have a loop variable that exists outside of the scope of the loop body, and in both cases that variable is being mutated.
The only way I can explain this output is if the let declared loop variable (block-scoped) is "magically" copied when it is captured.
Did I get this right?

Variable meaning / value in a comma separated syntax [duplicate]

This question already has answers here:
What does i = (i, ++i, 1) + 1; do?
(7 answers)
Closed 4 years ago.
Could you tell me what this i means after the comma. I am struggling to understand this syntax or even find it in the internet.
let i;
this.questions = document.querySelectorAll('.question'), i;
i is not used anymore in the function
If setting a debugger after this lines. i is undefined

It doesn't seem to be of any use indeed.
As an information, it's just to declare / instantiate your variables. Take a look :
let i, j;
j = 1, i = 2;
console.log(i, j);

JS, dynamic callback function (i.e. dynamic constants), click example [duplicate]

This question already has answers here:
JavaScript closure inside loops – simple practical example
(44 answers)
Closed 6 years ago.
My code is:
for (i = 0; i < 3; i++) {
elem[i].click(function(event) { alert(i); });
}
Unfortunately that does not work. Any idea how to make it alert the same i as the i in elem[i].
Thank you!

You should use let for defining variable i. Unlike var, variables defined by let are block scoped. As a result, they won't be overwritten after each iteration.
for (let i = 0; i < 3; i++) {
elem[i].click(function(event) { alert(i); });
}
By the way, when ES6 is available, it's generally a bad idea to use closures in this case. It's kind of overkill.

Explaining IIFE Inside Loop [duplicate]

This question already has answers here:
JavaScript closure inside loops – simple practical example
(44 answers)
How do JavaScript closures work?
(86 answers)
Closed 6 years ago.
I read this article, but I have some issues understanding it.
var funcs = [];
for (var i = 0; i < 3; i++) {
(function IIFE(arg) {
funcs[arg] = function output() {
console.log("My value: " + arg);
};
})(i);
}
for (var j = 0; j < 3; j++) {
funcs[j]();
}
How can I explain the mechanism of this snippet?
Start: i has value of 0. IIFE gets the value of i and stores it in arg.
funcs[0] just get assigned to a function named output.
second time, when i is 1, funcs[1] gets a totally new function? Is this false?
.....
The problem is here, How does funcs[0](); know that arg was 0? How things interact?
Function output() looks up and ask IIFE function to give the required value? right? Can someone say which illustration is correct?
By Providing this picture, for example, number 1, I mean, when funcs[0]() gets executed, output looks for arg, it doesn't find it, then goes to higher scope and finds it. each function has it's own IIFE, for example. I'n not sure how far are this from Javascript or even programming world, but help me get it!