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How do I load the values from one array, into another array, without changing the memory address of the array receiving the new values?

If I do this:
a = []
b = [1, 2, 3]
a = b
a will then refer to b
I want to copy all of the values in b into a without changing any memory address/references.

You could push the values without changing the reference from a or b.
var a = [],
b = [1, 2, 3];
a.push(...b);

If you want to populate a in-place, without ever discarding the initial array, then you can simply loop over b and add all the items in a:
var a = []
var b = [1, 2, 3]
var firstA = a; //get a initial element
b.forEach(function(item) {
this.push(item)
}, a);// <-- pass `a` as the `this` context
console.log("`a` is unchanged", a === firstA);
console.log("`a` is not `b`", a !== b);
console.log(a);

let a = b.map(x => x);
The .map() function will create a new array with values generated from the source elements via the passed-in callback function.
As noted in a comment
let a = b.slice();
is pretty much exactly the same.
Basically this saves you the step of
let a = [];
since one brand-new empty array is as good as another. However, if you really have your heart set on using the empty array explicitly initialized, you could use
b.forEach(x, i) { a[i] = x; });

A= b.slice(). It will create a shadow copy of element without changing original one. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/slice

Array as a Variable Assignment in Javascript

If:
var x = [1, 2, 3];
var y = [4, 5, 6];
var z = x;
and then if z[2] = y[0];
Why is it that console.log(x); is [1, 2, 4] and not [1, 2, 3]?

When you do var z = x; you are no creating a new array, entirely separate from x, you are simply creating a reference to the original array. Hence, the change happens in both.
If you want to create a new object, you can use the new ES6 spread operator
var z = {...x};
Have a look at this answer for a more in-depth explanation of passing by reference and by value.

Cuz the 'z' variable is a pointer to the same array that 'x' point to.

An array in JavaScript is also an object and variables only hold a reference to an object, not the object itself. Thus both variables have a reference to the same object. So changing made through one variable reflected in other as well.
var x = [1, 2, 3];
var y = [4, 5, 6];
var z = x;
z[2]=y[0];
console.log(x);
var w=Object.assign([],x);
w[0]=y[1];
console.log(x);
console.log(w);
Look at the example. If you want to change in a new variable and don't want reflected that change in original one than use Object.assign.

JavaScript object property gets modified in a very interesting way

I have put together a small test, can someone explain why does JavaScript do the following?
Why will testArr inherit the modifications of newArray?
var TestClass = function() {
this.testArr = [10,11];
this.addToArray = function() {
var newArray = this.testArr;
newArray.push(12);
}
}
var testClass = new TestClass();
testClass.addToArray();
console.log(testClass.testArr); // will be [10, 11, 12]

Because they're the same array. Variables contain references to arrays. You're assigning that reference to a new variable, and now both variables point to the same array.
Here's a much simpler test which reproduces the behavior:
x = [1, 2] // the value of x is a reference to an array
y = x // the value of y is a reference to the same array
y.push(3) // modify the array pointed to by both variables
console.log(x) // [1, 2, 3]
If you want to create a new array, you need to clone the array.

With following statement
var newArray = this.testArr;
You will not copy the values from inside testArr. You will create a reference to this array. When you make an adjustment to a reference you will automatically change the referenced source too.
When you want to copy the values from testArr to another variable you can do this instead.
var newArray = this.testArr.slice();
This will copy the values from testArr inside newArray.

Reference in slice in javascript does not exit.

I am getting back to JS after a very long break and as I was going over "Jump Start JavaScript" book. In part of it says:
if your array contains an array as one of its items, it will be
copied by reference. In other words, if the original array changes, so
will the copy.
So I tried to do in code but JS doesn't work! (correct me if I'm wrong)
Then in SOF I saw this thread:
Reference to slice of an array
Should I consider this as mistake/error in the book?!
Thanks,
AK

It's unfortunate that the phrase by reference gets thrown around like it does, because it has a perfectly clear meaning that is completely different from one the being (mis-) used here. All values in JavaScript are copied/passed by value. When dealing with objects, that value is a reference to the object. But it's still a value that is copied.
When you make a shallow copy of an array using slice, the elements that are references to objects have those same references copied to the new array. That is, the same position in both arrays will refer to the same object. So, yes, changing one changes the other because there is no other; they are the same object.
Proof:
var arr = [[7, 6, 5], 4, 5];
var copy = arr.slice();
arr[0][0] = 88;
console.log(copy[0][0]); // 88
Tangent on pass-by-reference
If you can't write a function like this:
function swap(arg1, arg2) {
var temp = arg1;
arg1 = arg2;
arg2 = temp;
}
...that actually swaps the values of two variables:
var one = [1, 2, 3];
var two = ["a", "b", "c"];
swap(one, two);
console.log(one) // same as before
...then your language does not support pass by reference.
More pass-by-value
Wait, doesn't the following demonstrate that JavaScript must be pass-by-reference:
function Test(arg1) {
this.arg = arg1;
}
var one = [1, 2, 3];
var x = new Test(one);
x.arg[0] = 5;
console.log(one); // it's been updated!
No! If you think this then you still haven't understood what pass by reference actually is. Yes, a reference to the original array was passed to Test, but it was passed by value. In other words, x.arg refers to the same object in memory as one but it is not an alias for one. To prove this, you simply need to assign something new to x.arg:
x.arg = ["new array"];
This does not affect one. These two variables are independent. JavaScript is strictly pass-by-value. There are no exceptions.

In JavaScript it´s also correct to initialize an Array with the "new" keyword:
var arr = new Array();
arr.push("0. Index");
arr.push("1. Index");
document.body.appendChild(document.createTextNode(JSON.stringify(arr)));
In every programming language I know, the new keyword stands for creating an object. And also in every programming language I know, an object is Call-By-Reference and not Call-By-Value. So the book tells you the truth. That´s a JavaScript special: Because PHP arrays are always Copy-By-Value, if you do not use explicit the & for making it to a reference.
To clear this this disussion finally, execute this code below in you browser and compare the used memory size. A 2nd reference to an already existing object should cause lower memory usage than 2 seperated arrays:
First this one:
var arr1 = new Array();
for (var i = 0; i < 11; i++)
{
arr1.push(i);
}
var arr2 = arr2;
Second this one:
var arr1 = new Array();
var arr2 = new Array();
for (var i = 0; i < 11; i++)
{
arr1.push(i);
arr2.push(i);
}
Or simply try this one and see the results:
var arr1 = new Array();
for (var i = 0; i < 11; i++)
{
arr1.push(i);
}
var arr2 = arr1;
arr2[5] = "test";
document.body.appendChild(document.createTextNode("arr2 => " + JSON.stringify(arr2)));
document.body.appendChild(document.createTextNode("arr1 => " + JSON.stringify(arr1)));
If the output prints on the 5th index for both arrays "test" instead of 5, it´s for sure Call-By-Reference, as your book said. If not, it´s Copy-By-Value.

var subarray = [1, 2, 3];
var main_array = ['a', subarray, 'b'];
var slice = main_array.slice(0, 2);
console.log(slice); => ['a', [1, 2, 3]]
subarray[0] = 10;
console.log(slice); => ['a', [10, 2, 3]]

JavaScript create use backup array

Just look at the code and you'll understand what I mean:
​var aBackup = [3, 4]; // backup array
​​var a = aBackup; // array to work with is set to backup array
a[0]--; // working with array..
a = aBackup; // array o work with will be rested
console.log(a); // returns [2, 4] but should return [3, 4]
console.log(aBackup);​ // returns [2, 4] too​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​ but should return [3, 4] too

You need to make real copies of your Arrays instead of just using a reference:
var aBackup = [3, 4]; // backup array
var a = aBackup.slice(0); // "clones" the current state of aBackup into a
a[0]--; // working with array..
a = aBackup.slice(0); // "clones" the current state of aBackup into a
console.log(a); // returns [3, 4]
console.log(aBackup); // returns [3, 4]
See MDN for documentation on the slice-method

Doesn't javascript uses pointer for arrays ? Should ​​var a = aBackup; do a copy ? otherwise the results seems normal to me...

An array is a reference type of object, so changes made to it will change the underlying value it points to, a and aBackup will point to the same value, and a change made to a will change aBackup aswell.

It is because when you do this, you are not making a copy of the array, but infact a reference to the original array.
var a IS aBackup; // if you will
When you need to do is clone the backup array.