We Keep Coding

Why do changes made to a cell propagate to other cells in this 2 dimensional array created using fill?

I have a problem with this 2-dimensional array in JS. When I change a[1][0], a[0][0] changes with it. Is there something wrong in the way I am initializing it? If yes, how can I initialize it properly?
>var a = Array(100).fill(Array(100).fill(false));
>a[0][0]
>false
>a[1][0]
>false
>a[1][0] = true
>true
>a[1][0]
>true
>a[0][0]
>true

var a = Array(100).fill(Array(100).fill(false));
a contains an array, each element of which references to an array. you are filling the outer array with an array which contains all false values. The inner array is being made only once and reference to the same array is passed to each element of outer array that is why if you perform an operation on one element it reflects on other elements as well.
This is actually equivalent to
var a1 = Array(100).fill(false);
var a = Array(100).fill(a1);
here a gets 100 elements all having reference to same array a1. So if you change one element of a, all elements change since they are references to same array.
you will need to fill each element in outer array with a new array. you can do something like this:
var a = [];
for(var i=0; i<100; i++)
a.push(Array(100).fill(false));

Your entire array will be filled with references to the same (second dimension) array object.
To fill it with distinct objects, you'd have to do something like this:
const a = Array(100).fill(false).map(x => Array(100).fill(false));
a[0][0] = true;
console.log(a[0][0]);
console.log(a[0][1]);
console.log(a[1][1]);
Note that the values in the initial array need to be explicitly set to something (in my example false, but could also be undefined), because the array created by the constructor is sparse and the map() function will only operate on properties that actually exist.
To work around that, you could use Array.from():
const a = Array.from(Array(100), x => Array(100).fill(false));
a[0][0] = true;
console.log(a[0][0]);
console.log(a[0][1]);
console.log(a[1][1]);

Your problem is that you are using the second Array(100).fill(false) in every position of the first array. Thats why when you change one value in the "second" dimension it will change in every single position of the array. If you want to avoid this you hve to create a new Array for each position of the initial.
var x = Array(100).fill().map(x => Array(100).fill(false));

what you are doing here is adding the same array object at each index of bigger array.
So when you do
Array(x).fill(Array(y).fill(false));
What you are doing actually is :
Array(x).fill(Y); // i.e. X.fill(Y)
Now whenever you change Y, you will get the same value at each index of X.
You should use loop to fill in data when data itself is an object(remember Array is object).

These looks like a duplicate question but I've always just used something like this
var x = new Array(10);
for (var i = 0; i < 10; i++) {
x[i] = new Array(10).fill(false);
}
x[5][5] = true
Credit: How can I create a two dimensional array in JavaScript?

Since arrays are reference types, creating an N Dimensional array in JS is not so trivial. You need to create a clone tool first.
Array.prototype.clone = function(){
return this.map(e => Array.isArray(e) ? e.clone() : e);
};
function arrayND(...n){
return n.reduceRight((p,c) => c = (new Array(c)).fill().map(e => Array.isArray(p) ? p.clone() : p ));
}
var arr = arrayND(2,3,4,"x")
console.log(JSON.stringify(arr));
arrayND takes indefinite number of integer arguments each designating the size of a dimension and the last argument is the fill value.

The fill function as described in mdn, copies the same value to all array indices. You can use the below utility to create clones of the array. But since i am using the slice operator, this is restricted to cloning arrays.
https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/fill
if (!Array.prototype.clonefill) {
Object.defineProperty(Array.prototype, 'clonefill', {
value: function(value) {
// Steps 1-2.
if (this == null) {
throw new TypeError('this is null or not defined');
}
var O = Object(this);
// Steps 3-5.
var len = O.length >>> 0;
// Steps 6-7.
var start = arguments[1];
var relativeStart = start >> 0;
// Step 8.
var k = relativeStart < 0 ?
Math.max(len + relativeStart, 0) :
Math.min(relativeStart, len);
// Steps 9-10.
var end = arguments[2];
var relativeEnd = end === undefined ?
len : end >> 0;
// Step 11.
var final = relativeEnd < 0 ?
Math.max(len + relativeEnd, 0) :
Math.min(relativeEnd, len);
// Step 12.
while (k < final) {
O[k] = value.slice(0);
k++;
}
// Step 13.
return O;
}
});
}
Array(100).clonefill(Array(100))
a[0][0] = true
a[1][0]
false

Javascript use original array when creating subarray

I am new to Javascript and I have encountered the following problem:
I have an array of objects that remains unchanged. From this array, I would like to create subarrays with different start and end indexes.
var origArr = [o1, o2, o3, o4, o5];
// origArr .length = 5
var subArr = origArr.slice(1, 4);
// subArr = [o2, o3, o4];
Normally I can use the slice method. But this means that the elements are copied into the new array. Since all I want to do is the alter the start and end index, is there any easy way to do this without creating a new copy each time? I would just use the reference to the original array.
Thank you for your help.
Chris

You can't create a "view" of a subset of an array with standard JavaScript arrays. You can do it with the new typed arrays, but the new typed arrays only offer storage for various numeric formats. Your variable names (o1 and so on) suggest object references instead.
For standard arrays, you can use a function instead:
function subMapper(array, index, length) {
return function(i, value) {
if (index + i >= length) {
// Beyond the end of the array, do what you think is appropriate
}
if (arguments.length === 1) {
// Getting a value
return array[index + i];
}
// Setting a value
array[index + i] = value;
};
}
var origArr = [o1, o2, o3, o4, o5];
var subArrMapper = subMapper(origArr, 1, 4);
var x = subMapper(0);
console.log(x === o2); // true

How to remove all undefined keys from a javascript array (hopefully when creating the array?)

I have placed my frustrations into a jsfiddle to observe here: http://jsfiddle.net/8ShFr/1/
var brand_new_array = new Array();
brand_new_array[10] = "random array value";
alert('why does this array have a length of ' + brand_new_array.length + '???');
I am doing some calculations client side that require me to set javascript array keys of 1M+ in number.
Not knowing exactly what that number is demands that I iterate through the first 1M+ empty array values before getting to an array key that holds data.
I simply want to set a single large key value for a javascript array without creating a bunch of empty keys before it?
I am using jQuery.each to iterate over the array, and it keeps going through array[0], array[1], array[2], etc... when I only set array[123125] for example.

Just filter out the undefineds.
brand_new_array = brand_new_array.filter(function(n){return n !== undefined});

The reason for the length being 10 is that an array's length is set to the largest index number in the array. However, this does not mean there are 9 other values in there because in javascript an array is at its base an object.
The length is just a property in the object. Arrays in javascript are at their core objects (Array Object 1). They merely act like arrays through an api.
"Whenever a property is added whose name is an array index, the length property is changed, if necessary, to be one more than the numeric value of that array index" 1
1. ECMAScript Language Specification 15.4 Array Objects

You probably want to just use an object with strings for keys (the keys can be the toString() of Numbers, which will happen automatically if you try to use numbers).
var sparse_array_obj = {};
sparse_array_obj[10003210234] = 4; // Fair dice roll
sparse_array_obj[5] = 17; // Truly random number
sparse_array_obj[900] = Math.random(); // Pseudorandom number
for(var i in sparse_array_obj)
console.log(sparse_array_obj[i]);
The downside is that Javascript provides no guarantees about the iteration order through an object (since its keys are unordered by definition). There are however ways around this, such as:
// Sort the keys in numeric order
var sorted_keys = Object.keys(sparse_array_obj).sort(function(a, b){ return a - b; });
for(var i = 0; i < sorted_keys.length; i++)
console.log(sparse_array_obj[sorted_keys[i]]);
Object.keys needs to be shimmed in older browsers.

var brand_new_array = new Array();
brand_new_array[10] = "random array value";
var result = brand_new_array.filter(function(e) { return e != undefined;})[0];
alert(brand_new_array.indexOf(result));

Travis J is right. The array in your example only contains one entry, but your use of jQuery.each() is making you think there are 10 entries because it iterates from 0 up to the highest index number of the array (defines the length). This is from the jQuery.each() API documentation.
A generic iterator function, which can be used to seamlessly iterate over both objects and arrays. Arrays and array-like objects with a length property (such as a function's arguments object) are iterated by numeric index, from 0 to length-1. Other objects are iterated via their named properties.
Going back to your example:
var brand_new_array = new Array();
brand_new_array[10] = "random array value";
This will result in only one console.log output:
for(var i in brand_new_array)
console.log(brand_new_array[i]);
This will result in 10 console.log outputs:
$(brand_new_array).each( function(i,e) { console.log(e) })
Similarly, this will result in 10 console.log outputs:
for (var i=0;i<brand_new_array.length;i++)
console.log(brand_new_array[i]);
If you really want to stick with using .each() then you can skip the undefined indices like so:
$(brand_new_array).each( function(i,e) {
if (this.hasOwnProperty(i)){ console.log(e) }
})

Filter the falsy items - including undifined:
var a=[1,2,"b",0,{},"",NaN,3,undefined,null,5];
var b=a.filter(Boolean); // [1,2,"b",{},3,5]

The length is 11 because the index starts at 0.
x[0] = undefined
x[1] = undefined
x[2] = undefined
x[3] = undefined
x[4] = undefined
x[5] = undefined
x[6] = undefined
x[7] = undefined
x[8] = undefined
x[9] = undefined
x[10] = "random array value"

Difference between associative [],{} and object in javascript [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What is the difference between an array and an object?
The item exists in the array but it says that the array is 0 length?
I m a bit confused in object and associative array in javascript. I read this :question but this question says that there is not much of a difference in both. I wrote this in the console:
var a = [];
a["A"] = 1;
var b = {};
b["B"] = 2;
var c = new Object();
c["C"] = 3;
output for above are as:
a gives {A : 1}
b gives {B : 2}
c gives {C : 3}
All above three cases gives same reault as in they all gives an object. Question is how all above 3 cases are treated in javascript.

An array is also an object, that's why you can use non-numeric indices also. Those will be added as properties to the array object, not part of the array data.
The difference between an array and an object, is that the array counts properties with a numeric index as being part of the array data, and updates the length property accordingly. (Also the Array object has some specific methods to work with the array data.)
var a = [];
a[42] = 1337;
Now the length has changed to include the item:
alert(a.length); // shows 43
Using strings or numbers as index doesn't matter, a numeric index counts even if it's a string:
alert(a[42]); // shows 1337
alert(a["42"]); // shows 1337
Reducing the length removes the properties outside it:
a.length = 10;
alert(a[42]); // shows undefined

In your example, b and c are essentially the same thing, because {} is the equivalent of new Object().
Coming back to a, it's defined as an Array which is a special kind of Object in the sense that numeric properties (based on uint32) are treated differently. Its length property gets updated when those properties are added and removed.
When you use 'A' as an index, it gets treated as a regular property, defined by how Object works with properties, so you can access it as a.A or a['A'] whereas an index of [5] can only be accessed using a[5].
Normally, the debug output of an array is always [] unless the length property is non-zero, so the output you've shown is somewhat irregular.
Trivia
According to the ECMAScript documentation, a particular value p can only be an array index if and only if:
(p >>> 0 === p) && (p >>> 0 !== Math.pow(2, 32) - 1)
See also:
The item exists in the array but it says that the array is 0 length?

Let's start by clarifying something:
new Object() is the same as {}
new Array() is the same as []
The latter are just shortened forms of the former.
Behind the scenes, everything in javascript is basically an object (this is an exaggeration but fairly accurate). Arrays are simply derived from objects. Here's a fairly rudimentary example of what an Array REALLY looks like:
var myArray = {};
myArray[0] = 'value1';
myArray[1] = 'value2';
myArray[2] = 'value3';
myArray[length] = 3;
The prototype of an array contains all the methods. For example:
// myArray#push
myArray.prototype.push = function(val){
myArray[this.length] = val;
this.length++;
}
Another way to illustrate this is to take an array and add keys that are not numeric:
var example = [];
example.push(0);
example.push(1);
example.push(2);
example['a'] = 'a';
example['b'] = 'b';
example['c'] = 'c';
example.log = function(){
for(var i = 0, l = this.length; i < l; i++){
console.log( example[i] );
}
console.log(this['a']);
console.log(this['b']);
console.log(this['c']);
}
// example[0] is 0
// example[1] is 1
// example[2] is 2
// example.log is my custom function
example.log(); // results in
// 0
// 1
// 2
// a
// b
// c
Also, don't always believe everything the console tells you. For example, if you do this:
var console_confusion = {};
console_confusion.length = 100;
console_confusion.splice = function(){ /* do absolutely nothing */ };
console.log( console_confusion );//results in
//
// in Chrome:
// [undefined × 100]
//
Chrome will interprut anything with a numeric length property and a splice function as an Array. This is why jQuery objects look like Arrays in the console.

Please read the following article – http://www.2ality.com/2012/12/arrays.html
In short, “regular arrays”, denoted as [] in JavaScript, are also objects, just like {}, but they have length property and “numeric” keys (“indicies”), plus their internal __proto__ property points at Array.prototype object, which holds all Array methods, such as push or forEach etc.

first is an array
second is an object array
third is an array object

Difference between Array.length = 0 and Array =[]?

Can some one explain the conceptual difference between both of them. Read somewhere that the second one creates a new array by destroying all references to the existing array and the .length=0 just empties the array. But it didn't work in my case
//Declaration
var arr = new Array();
The below one is the looping code that executes again and again.
$("#dummy").load("something.php",function(){
arr.length =0;// expected to empty the array
$("div").each(function(){
arr = arr + $(this).html();
});
});
But if I replace the code with arr =[] in place of arr.length=0 it works fine. Can anyone explain what's happening here.

foo = [] creates a new array and assigns a reference to it to a variable. Any other references are unaffected and still point to the original array.
foo.length = 0 modifies the array itself. If you access it via a different variable, then you still get the modified array.
Read somewhere that the second one creates a new array by destroying all references to the existing array
That is backwards. It creates a new array and doesn't destroy other references.
var foo = [1,2,3];
var bar = [1,2,3];
var foo2 = foo;
var bar2 = bar;
foo = [];
bar.length = 0;
console.log(foo, bar, foo2, bar2);
gives:
[] [] [1, 2, 3] []
arr.length =0;// expected to empty the array
and it does empty the array, at least the first time. After the first time you do this:
arr = arr + $(this).html();
… which overwrites the array with a string.
The length property of a string is read-only, so assigning 0 to it has no effect.

The difference here is best demonstrated in the following example:
var arrayA = [1,2,3,4,5];
function clearUsingLength (ar) {
ar.length = 0;
}
function clearByOverwriting(ar) {
ar = [];
}
alert("Original Length: " + arrayA.length);
clearByOverwriting(arrayA);
alert("After Overwriting: " + arrayA.length);
clearUsingLength(arrayA);
alert("After Using Length: " + arrayA.length);
Of which a live demo can be seen here: http://www.jsfiddle.net/8Yn7e/
When you set a variable that points to an existing array to point to a new array, all you are doing is breaking the link the variable has to that original array.
When you use array.length = 0 (and other methods like array.splice(0, array.length) for instance), you are actually emptying the original array.

Are you sure it really works?
I did a little experiment here, and trying to "add" an Array with a String resulted in a string.
function xyz(){
var a = [];
alert(typeof(a+$("#first").html()));
// shows "string"
}
http://www.jsfiddle.net/4nKCF/
(tested in Opera 11)