CODEWITHSUNDEEP
javascript
user1637281
user1637281

Reputation:

What happens to properties on a constructor function?

Suppose there is:

function someConstructor(){
   // initialize
}
someConstructor.prop = {test:'test'};

var obj1 = new someConstructor();
var obj2 = new someConstructor();

My guess would be that there is still only one copy of prop available via someConstructor.prop for obj1 and obj2 available via someConstructor.

Upvotes: 0

Views: 64

Answers (2)

Bergi
Bergi

Reputation: 664206

function SomeConstructor(){}; someConstructor.prop = …

JavaScript is case sensitive, this is example won't work.

My guess would be that there is still only one copy of prop for obj1 and obj2.

No. There won't be any copies at all. prop is a property of the function object, and has nothing to do with obj1 or obj2. It is not inherited by them.

And that's the difference to using a property on the prototype, which is the object from which obj1 and obj2 do inherit.

a static property that is copied directly to the object

Only if by "object" you mean the constructor function here. In JS, there are no classes, and no "static" properties. Yet, the correspondent of "static class attributes" would be properties on the constructor function.

Upvotes: 0

Shadow Man
Shadow Man

Reputation: 3402

Prototype is used to make sure that 1 and only 1 object is there for all instances. However those instances may override that prototype with their own objects. In firefox at least, setting Test.opts does not allow o1 or o2 to inherit it. But setting Test.prototype.opts does.

Try the following:

var Test = function () {};
Test.opts = {helo: "hello"};

var o1 = new Test();
var o2 = new Test();
console.log( o1.opts ); // undefined
o1.opts = {helo: "bye"};
console.log( o2.opts ); // undefined

Test.prototype.opts = {helo: "hello"};
console.log( o1.opts ); // Object { helo: "bye" }
console.log( o2.opts ); // Object { helo: "hello" }

Upvotes: 4

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