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Seems like the following code should return a true, but it returns false.
var a = {};
var b = {};
console.log(a==b); //returns false
console.log(a===b); //returns false
How does this make sense?
The only difference between regular (==) and strict (===) equality is that the strict equality operator disables type conversion. Since you're already comparing two variables of the same type, the kind of equality operator you use doesn't matter.
Regardless of whether you use regular or strict equality, object comparisons only evaluate to true if you compare the same exact object.
That is, given var a = {}, b = a, c = {};, a == a, a == b, but a != c.
Two different objects (even if they both have zero or the same exact properties) will never compare equally. If you need to compare the equality of two object's properties, this question has very helpful answers.
How does this make sense?
Because "equality" of object references, in terms of the == and === operators, is purely based on whether the references refer to the same object. This is clearly laid out in the abstract equality comparison algorithm (used by ==) and the strict equality comparison algorithm (used by ===).
In your code, when you say a==b or a===b, you're not comparing the objects, you're comparing the references in a and b to see if they refer to the same object. This is just how JavaScript is defined, and in line with how equality operators in many (but not all) other languages are defined (Java, C# [unless the operator is overridden, as it is for string], and C++ for instance).
JavaScript has no inbuilt concept of equivalence, a comparison between objects that indicates whether they're equivalent (e.g., have the same properties with the same values, like Java's Object#equals). You can define one within your own codebase, but there's nothing intrinsic that defines it.
As from The Definitive Guide to Javascript.
Objects are not compared by value: two objects are not equal even if they have the same properties and values. This is true of arrays too: even if they have the same values in the same order.
var o = {x:1}, p = {x:1}; // Two objects with the same properties
o === p // => false: distinct objects are never equal
var a = [], b = []; // Two distinct, empty arrays
a === b // => false: distinct arrays are never equal
Objects are sometimes called reference types to distinguish them from JavaScript’s primitive types. Using this terminology, object values are references, and we say that objects are compared by reference: two object values are the same if and only if they refer to the same underlying object.
var a = {}; // The variable a refers to an empty object.
var b = a; // Now b refers to the same object.
b.property = 1; // Mutate the object referred to by variable b.
a.property // => 1: the change is also visible through variable a.
a === b // => true: a and b refer to the same object, so they are equal.
If we want to compare two distinct objects we must compare their properties.
use JSON.stringify(objname);
var a = {name : "name1"};
var b = {name : "name1"};
var c = JSON.stringify(a);
var d = JSON.stringify(b);
c==d;
//true
Here is a quick explanation of why {} === {} returns false in JavaScript:
From MDN Web Docs - Working with objects: Comparing objects.
In JavaScript, objects are a reference type. Two distinct objects are never equal, even if they have the same properties. Only comparing the same object reference with itself yields true.
// Two variables, two distinct objects with the same properties
var fruit = {name: 'apple'};
var fruitbear = {name: 'apple'};
fruit == fruitbear; // return false
fruit === fruitbear; // return false
// Two variables, a single object
var fruit = {name: 'apple'};
var fruitbear = fruit; // Assign fruit object reference to fruitbear
// Here fruit and fruitbear are pointing to same object
fruit == fruitbear; // return true
fruit === fruitbear; // return true
fruit.name = 'grape';
console.log(fruitbear); // output: { name: "grape" }, instead of { name: "apple" }
For more information about comparison operators, see Comparison operators.
How does this make sense?
Imagine these two objects:
var a = { someVar: 5 }
var b = { another: 'hi' }
Now if you did a === b, you would intuitively think it should be false (which is correct). But do you think it is false because the objects contain different keys, or because they are different objects? Next imagine removing the keys from each object:
delete a.someVar
delete b.another
Both are now empty objects, but the equality check will still be exactly the same, because you are still comparing whether or not a and b are the same object (not whether they contain the same keys and values).
===, the strictly equal operator for objects checks for identity.
Two objects are strictly equal if they refer to the same Object.
Those are two different objects, so they differ.
Think of two empty pages of paper. Their attributes are the same, yet they are not the same thing. If you write something on one of them, the other wouldn't change.
This is a workaround: Object.toJSON(obj1) == Object.toJSON(obj2)
By converting to string, comprasion will basically be in strings
In Javascript each object is unique hence {} == {} or {} === {} returns false. In other words Javascript compares objects by identity, not by value.
Double equal to ( == ) Ex: '1' == 1 returns true because type is excluded
Triple equal to ( === ) Ex: '1' === 1 returns false compares strictly, checks for type even
This question already has answers here:
Which equals operator (== vs ===) should be used in JavaScript comparisons?
(48 answers)
Closed 1 year ago.
The community reviewed whether to reopen this question 1 year ago and left it closed:
Original close reason(s) were not resolved
const abc = 5;
const bcd = 5
console.log(abc===bcd)/// return true
This result is surprising to me.
Since abc will be assigned a different memory location ...and bcd has also a different location
why this is returning true
If the code was
const abc = 5
const bcd = abc
console.log(abc === bcd) ///true here make sense
I seem to be be really confused with the first case since abc and bcd are two different variables and have no relation between them
Any related article or blog would really help.
In javascript all primitives (string, nubmer, bigint, boolean, undefined, symbol, null) are immutable and will be compared by value:
All primitives are immutable, i.e., they cannot be altered. It is important not to confuse a primitive itself with a variable assigned a primitive value. The variable may be reassigned a new value, but the existing value can not be changed in the ways that objects, arrays, and functions can be altered.
In comparison to that objects (this also includes arrays and functions, basically everything that is not a primitive) are mutable and will be compared by identity.
Example:
console.log("Primitives compare by value:");
console.log(5 === 5); // true
console.log("foo" === "foo"); // true
console.log(true === true); // true
console.log("Objects compare by identity:");
console.log({} === {}); // false
console.log([] === []); // false
console.log(function(){} === function(){}); // false
Primitive Wrappers
Javascript also has wrapper objects for the primitive types, which might be the source of your question.
These wrappers wrap a primitive value, and are - as the name suggests - objects. So for primitive wrapper instances your code would be correct:
let a = new Number(1);
let b = new Number(1);
console.log("Primitive wrappers are objects:");
console.log(a === a); // true
console.log(a === b); // false
There is a fundamental difference in JavaScript between primitive values (undefined,null, booleans, numbers, and strings) and objects (including arrays and functions)
I. Primitives are compared by value:
Two values are the same only if they have the same value.
If two distinct string values are compared, JavaScript treats them as equal if,
and only if, they have the same length and if the character at each index is the same.
II. Objects are different than primitives.
Objects are not compared by value: Two distinct objects are not equal even if they have the same properties and values.
Therefore, Objects are sometimes called reference types to distinguish them from JavaScript’s primitive types.
Seems like the following code should return a true, but it returns false.
var a = {};
var b = {};
console.log(a==b); //returns false
console.log(a===b); //returns false
How does this make sense?
The only difference between regular (==) and strict (===) equality is that the strict equality operator disables type conversion. Since you're already comparing two variables of the same type, the kind of equality operator you use doesn't matter.
Regardless of whether you use regular or strict equality, object comparisons only evaluate to true if you compare the same exact object.
That is, given var a = {}, b = a, c = {};, a == a, a == b, but a != c.
Two different objects (even if they both have zero or the same exact properties) will never compare equally. If you need to compare the equality of two object's properties, this question has very helpful answers.
How does this make sense?
Because "equality" of object references, in terms of the == and === operators, is purely based on whether the references refer to the same object. This is clearly laid out in the abstract equality comparison algorithm (used by ==) and the strict equality comparison algorithm (used by ===).
In your code, when you say a==b or a===b, you're not comparing the objects, you're comparing the references in a and b to see if they refer to the same object. This is just how JavaScript is defined, and in line with how equality operators in many (but not all) other languages are defined (Java, C# [unless the operator is overridden, as it is for string], and C++ for instance).
JavaScript has no inbuilt concept of equivalence, a comparison between objects that indicates whether they're equivalent (e.g., have the same properties with the same values, like Java's Object#equals). You can define one within your own codebase, but there's nothing intrinsic that defines it.
As from The Definitive Guide to Javascript.
Objects are not compared by value: two objects are not equal even if they have the same properties and values. This is true of arrays too: even if they have the same values in the same order.
var o = {x:1}, p = {x:1}; // Two objects with the same properties
o === p // => false: distinct objects are never equal
var a = [], b = []; // Two distinct, empty arrays
a === b // => false: distinct arrays are never equal
Objects are sometimes called reference types to distinguish them from JavaScript’s primitive types. Using this terminology, object values are references, and we say that objects are compared by reference: two object values are the same if and only if they refer to the same underlying object.
var a = {}; // The variable a refers to an empty object.
var b = a; // Now b refers to the same object.
b.property = 1; // Mutate the object referred to by variable b.
a.property // => 1: the change is also visible through variable a.
a === b // => true: a and b refer to the same object, so they are equal.
If we want to compare two distinct objects we must compare their properties.
use JSON.stringify(objname);
var a = {name : "name1"};
var b = {name : "name1"};
var c = JSON.stringify(a);
var d = JSON.stringify(b);
c==d;
//true
Here is a quick explanation of why {} === {} returns false in JavaScript:
From MDN Web Docs - Working with objects: Comparing objects.
In JavaScript, objects are a reference type. Two distinct objects are never equal, even if they have the same properties. Only comparing the same object reference with itself yields true.
// Two variables, two distinct objects with the same properties
var fruit = {name: 'apple'};
var fruitbear = {name: 'apple'};
fruit == fruitbear; // return false
fruit === fruitbear; // return false
// Two variables, a single object
var fruit = {name: 'apple'};
var fruitbear = fruit; // Assign fruit object reference to fruitbear
// Here fruit and fruitbear are pointing to same object
fruit == fruitbear; // return true
fruit === fruitbear; // return true
fruit.name = 'grape';
console.log(fruitbear); // output: { name: "grape" }, instead of { name: "apple" }
For more information about comparison operators, see Comparison operators.
How does this make sense?
Imagine these two objects:
var a = { someVar: 5 }
var b = { another: 'hi' }
Now if you did a === b, you would intuitively think it should be false (which is correct). But do you think it is false because the objects contain different keys, or because they are different objects? Next imagine removing the keys from each object:
delete a.someVar
delete b.another
Both are now empty objects, but the equality check will still be exactly the same, because you are still comparing whether or not a and b are the same object (not whether they contain the same keys and values).
===, the strictly equal operator for objects checks for identity.
Two objects are strictly equal if they refer to the same Object.
Those are two different objects, so they differ.
Think of two empty pages of paper. Their attributes are the same, yet they are not the same thing. If you write something on one of them, the other wouldn't change.
This is a workaround: Object.toJSON(obj1) == Object.toJSON(obj2)
By converting to string, comprasion will basically be in strings
In Javascript each object is unique hence {} == {} or {} === {} returns false. In other words Javascript compares objects by identity, not by value.
Double equal to ( == ) Ex: '1' == 1 returns true because type is excluded
Triple equal to ( === ) Ex: '1' === 1 returns false compares strictly, checks for type even
This question already has answers here:
Why doesn't equality check work with arrays [duplicate]
(6 answers)
Closed 7 years ago.
I am storing (x,y) coordinates as 2-element arrays.
var coordinateA = [0,3];
var coordinateB = [1,2];
I also have a longer array containing many of these coordinates:
var coordinates = [coordinateA, coordinateB]
Imagine my surprise when the following statements turned out to be false:
jQuery.inArray(coordinateA, coordinates); // returns -1
coordinateA == coordinates[0]; // returns false
[0,3] == [0,3]; // returns false(!)
coordinateA == coordinateA; // returns true, thankfully
Could someone help me understand why this is the case? Also, is there a better way to represent 2D coordinates in Javascript? Thanks for any clues or suggestions.
This is because you have two separate array references.
The equality operator is checking that the references are equal, not the content of the arrays.
One of the puzzling things about JavaScript is how equality is dealt with. I will do my best to explain this.
The equality rules can be quite hard to grasp. Generally speaking, you can compare by relative equality (==) or strict equality (===).
relative equality:
This compares by value only and does not care about type.
Example
var x = '2';
var y = 2;
x == y;
=> false;
In relative equality, the string "2" equals the number 2. This will return true since types are not compared
strict equality
This compares by both value and type.
Example
var x = '2';
var y = 2;
x === y;
=> false
In this case, the string "2" does NOT equal the number 2. Because String and Number are two different types.
Comparisons with arrays and objects are done differently though.
In your case, arrays are considered objects.
typeof([1,2])
=> "object"
In JavaScript, all objects are different. They are compared by their object ids. To determine if arrays are equal, you have to perform type conversion to a string.
String([1,2]) == String([1,2])
=> true
However, the underscore library has an is_equal method that can determine whether two arrays are equal
_.isEqual(array1, array2);
Underscore does this by performing a deep comparison between two objects to determine if they should be considered equal.
It's important to note that order matters here, as it does in the string comparison.
_isEqual([1,2], [1,2])
=> true
_isEqual([1,2], [2,1])
=> false
I know that identical objects are not equal, i.e:
var obj = { name: "Value" };
var obj2 = { name: "Value" };
console.log("obj equals obj2: " + (obj === obj2)); //evaluates to false
Yet primitive types are:
var str = "string1";
var str2 = "string1";
console.log("str equals str2: " + (str === str2)); //evaluates to true
My question is why. Why are objects and primitives treated differently? If an object is nothing but an empty container, with only the attributes you specify to put in the container, why wouldn't the container's identical attributes evaluate to be the same? I looked around for this answer on SO and elsewhere, but didn't find an answer.
Is a JS object treated as something different in the DOM than a primitive type?
Thanks
This seems to really be a question about === so let's look at the Strict Equality Comparison Algorithm, in which point 7 says
Return true if x and y refer to the same object. Otherwise, return false.
So what does it mean to be "the same object"? It means they don't just look like eachother, but are at the same place in memory too. This means that the only time when an Object is === to an Object is when they're the same thing.
var a = {},
b = {}, // identical to `a`
c = a; // same as `a`
a === b; // false
a === c; // true
b === c; // false
When a variable's value is an object, well, it isn't an object: it's a reference to an object. Two variables that contain references to the same object are indeed equal:
var myObj = { hello: "world" };
var a = myObj;
var b = myObj;
if (a == b) alert("YES!!"); // YES!!
When the == operator has object references on both sides, the comparison made is to test whether the objects refer to the same object. When primitive values are involved, the semantics are different: the values are directly compared.
Generally, === operator checks for types, and if they are the same, checks values. Object type contains a reference, so, to be equal, they have to reference the same object and be of the same type. String literal value is not a reference, it is a value, so the === will produce true for string literals, but not for "abc" === new String("abc") because latter is an Object.
More information can be found here: A lot of details can be explored from here: Which equals operator (== vs ===) should be used in JavaScript comparisons?
First off, JavaScript objects aren't part of the DOM. The DOM (Document Object Model) are the HTML elements which make up your page. They cooperate together, but aren't directly linked.
Basically, yes, primitives are a special case. You can kind of think of it as if the value of a primitive is a constant (in a sense).
For example, take the example of the number 5. No matter how many times I declare 5, 5 will always equal 5. Thus, it isn't a stretch to say that {var a holding the value 5} is equivalent to {var b holding the value 5}. This concept is a little fuzzier with strings, but it still holds. A string that is "abc" is always the same as any other variable holding a string that is "abc".
This doesn't apply to objects either.
If you have two variables hold the same object, they are eqivalent.
var a = {};
var b = a;
console.log(a == b); // true
console.log(a === b); // true
However, if we create two objects that look similar:
var a = {};
var b = {};
console.log(a == b); // false
console.log(a === b); // false
This seems a bit weird at first, but think about the inner workings that are going on. Consider that when you pass an object in to a function, if you change that object, it is changed outside of the function to. It's passed by reference.
This means you can think of a pointer (a memory address) being stored in the variables. So, if you imagine that they have memory address in them (like 0x123456 and 0x654321), then it makes a little more sense (0x123456 and 0x654321 are different, so you wouldn't expend them to be equal). They are two separate things taking up their own area in the memory.
Make sense?
You can answer to this question at several levels.
strings
Factually, yes, strings are handled differently from objects as far as strict comparison operator is concerned.
Semantically, that is more convenient than having to resort to strcmp or equivalent mechanisms to compare two strings.
Implementation-wise, the cost is neglectible, so JavaScript can offer you that convenience.
By the way, people telling the strict equality operator checks if both variables point to the same memory location are wrong. In case of strings, === will succeed if the string contents are equal, wherever they might be located in memory.
Objects
Semantically, contrary to primitive types like numbers or strings, it is difficult to offer a consistent set of comparison operators for objects.
You could do an in-depth comparison for equality, but greater/lower operators would make little sense.
The choice of Javascript is rather inconsistent here.
the semantics of equality comparison (be it == or ===) are limited to references
(i.e. == or === will succeed if the references are equal).
Implementation-wise, a deep comparison could be quite costly.
There are also subtelties as how to interpret undefined properties.
At any rate, JavaScript did not choose to implement a deep comparison, so if you want one, you'll have to do it yourself.
And there have been terabytes of code written to try and provide the ideal in-depth object comparison function.
ordered comparison is handled quite differently.
You can define a valueOf method that will return whatever primitive value you want to be used for ordered comparison, e.g
myObject.prototype.valueOf = function(){return this.my_comparison_value; };
If not explicitely defined, valueOf will default to "[object Object]".
So if you don't supply a valueOf method:
< and > operators will always return false (which kind of makes sense).
>= and <= will always return true, regardless of the references being equal or not
(which makes a lot less sense).
Now if you take the pain to define a valueOf, equality comparison will still not use it.
The only way to have a consistent behaviour would be to combine <= and >=, e.g.
if (a >= b && a <= b) { // equality using valueOf
For browser-supplied primitive objects like DOM elements, the behaviour of ordering operators depends on what the browser decided to return as a default value.
I would not recomend using that unless you really know what you're doing.