If you're looking for documentation, I'm not sure where to find it. However, let me explain the result you received:

Local > Global

When you declare a global variable, it can be accessed anywhere in your file. In the "test()" function, when you write:

var a = b = 2;

You are creating a new variable 'a' that takes the value of the global variable 'b' and simultaneously changes the value of 'b' to 2 - effectively overriding its value.

Similarly, when you write (inside 'test()'):

var a, b; 

or

var a; 
var b;

You are declaring two variables that are local to your function. When you assign a value to 'b', due to the principle of local being greater than global, you may encounter two scenarios:

1. If you console.log(b) inside 'test()', you will get 2.
2. If you console.log(b) outside 'test()', you will get 44.

Declaration != Assignment

Remember:

• 'var a, b;' is a declaration.

• 'a = b = 25;' is an assignment (essentially a double assignment).

• 'var a = b = 25' combines declaration and assignment into one step.

I hope this explanation helps! Feel free to ask if anything is unclear or if you need further clarification. :)

let within a function's scope does not overwrite variables declared in an outer scope. Allow me to clarify the functionality of your exam() method:

// these are global variables:
let x = "out";
let y = "out";
let z = "out";
let u = "out";
let v = "out";
let w = "out";


function exam() {
    // the next line is equivalent to
    // let x; let y; z = "in";
    let x, y, z = "in"; 
    // the local variable y is set to "in"
    y = "in";
    // this translates to:
    // declare a local variable u which points to global v which points to global w
    // and assign them all the value "in"; hence only global v and global w get updated to "in";
   // *edited version: w = "in"; v = w; let u = v;
    let u = v = w = "in";
    // declare a local variable z and give it the value "in"
    let z = "in";
}
exam();
// when back in the global scope, x, y, z, u and v remain unchanged
// thus x == "out", y == "out", z == "out", u == "out", and v == "out"
// however, w == "in" and v == "in" due to modifications made by the exam() method

It's important to remember

When you assign primitive values like this:

var a = b = 2;

It is the same as:

var b = 2;
var a = b;

Both a and b will have the value of 2.

However, if you assign objects instead of primitive values:

var a = b = [1,2,3,4];

This means:

var b = [1,2,3,4];
var a = b;

This indicates that both a and b are referencing the same object. Therefore, any changes made to b will also affect a, and vice versa:

a.push(5); 
// a <--- [1,2,3,4,5] 
// b <--- [1,2,3,4,5] !! Be cautious! Changes made to b affect a too

Remember: Using a shorthand a = b = c = value assigns the same value to all variables. But when dealing with object assignment, they share the same reference pointing to a value. Always keep this in mind.

So for object assignment, this statement:

var a = b = [1,2,3,4]; // changing a WILL impact b

does not create the same effect as

var a = [1,2,3,4]; var b = [1,2,3,4]; // altering a won't affect b

One reason for this issue is that when you define variables within a function, they may not be accessible to $(".resultg").text("g = " + g);

However, in the line var d = e = f = "in";, a new local variable "d" is created which is set equal to global variables e and f, with the value "in". Since e and f are global, only in this instance will the reassignment to "in" be picked up by $(".resultg").text("g = " + g);

The question may not seem strange, and the answer is actually quite simple. It all boils down to understanding the concept of global variables.

// The following a,b,c,d,e,f,g are global variables

var a = "out";
var b = "out";
var c = "out";
var d = "out";
var e = "out";
var f = "out";
var g = "out";

---

Now, a,b,c,d,e,f,g have been assigned the value 'out'

If you attempt to alert any of them, it will display as 'out'

// Your function starts here

function test() {
    var a, b, c = "in"; // Here a,b,c are local variables and do not override the global variable values
    so this statement becomes meaningless when trying to print values from outside this function

    b = "in"; // Here lies the trick and confusion begins as b is actually a local variable created in the test function, not global
    Therefore, changing its value here will not affect the global one

    var d = e = f = "in"; // The same scenario applies here with e and f being global while d is local
    If we were to print the global variable values, we get:
    a=out
    b=out
    c=out
    d=out
    e=in
    f=in
    g=out

    var g = "in";

    Since this is also a local variable, the result will be:
    a=out
    b=out
    c=out
    d=out
    e=in
    f=in
    g=out
}

test();

$(".resulta").text("a = " + a);
$(".resultb").text("b = " + b);
$(".resultc").text("c = " + c);
$(".resultd").text("d = " + d);
$(".resulte").text("e = " + e);
$(".resultf").text("f = " + f);
$(".resultg").text("g = " + g);