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How can you define how objects are compared in JavaScript

I want to define a JavaScript class, Foo.

Foo = function(value){
    this.value = value;
};

I will create "instances" of my Foo:

foo1 = new Foo(1);
foo2 = new Foo(1);

and I want my instances of Foo to be comparable with each other using the standard == equality operator:

foo1 == foo2;      // this should be true

I can not find a way to do this. I thought I was on to something with the valueOf() function, but this is only useful when one side of the comparison is a primitive, not as above where both are of type object.

Have I missed something really simple akin to Ruby's

def ==(obj); end
like image 952
Chris Farmiloe Avatar asked Jan 13 '09 22:01

Chris Farmiloe


2 Answers

JavaScript does not have operator overloading. See this discussion covering your question.

like image 143
Crescent Fresh Avatar answered Nov 12 '22 04:11

Crescent Fresh


Not in a clean way...

Foo = function (value){
    this.value = value;
};

Override the toString function of your own object:

Foo.prototype.toString = function( ){ return this.value.toString(); }

Creating two test objects:

foo1 = new Foo(1);
foo2 = new Foo(1);

If the values of your objects is strings or numbers, you can have the javascript engine to convert your objects to a string by adding them to an empty string:

alert( ""+foo1 === ""+foo2 ); //Works for strings and numbers

The same thing but cleaner:

alert( foo1.toString() === foo2.toString() ); //Works for strings and numbers

If the values of your objects is numeric only, you can use the unary + operator to convert the object to a number:

alert( +foo1 === +foo2 ); //Works for numbers

However, I recommend you to both define the toString as above and also an equals:

Foo.prototype.equals=function(b){return this.toString() === b.toString();}

And then call it like this:

alert ( foo1.equals(foo2) );

Since you now have defined toString you can:

alert(foo1); // Alerts the value of foo1 instead of "[Object object]"
alert("foo1: " + foo1); // Alerts "foo1: 1". Useful when debugging. 
like image 22
some Avatar answered Nov 12 '22 03:11

some