Why can't I enumerate object property defined as not enumerable?

Question

I want to get the object properties that were defined via Object.defineProperty method.

Object.defineProperty(obj, prop, descriptor)

The Object.defineProperty() method defines a new property directly on an object, or modifies an existing property on an object, and returns the object.

• obj - The object on which to define the property.
• prop - The name of the property to be defined or modified.
• descriptor : The descriptor for the property being defined or modified.

So, let's take an example:

> a = {}
{}
> a.foo = "bar"
'bar'
> Object.keys(a)
[ 'foo' ]
> Object.defineProperty(a, "bar", { get: function () { return "foo"; }})
{ foo: 'bar' }
> a.bar
'foo'
> a.foo
'bar'
> Object.keys(a)
[ 'foo' ]
> for (k in a) { console.log(k); }
foo

In the for loop thing, how can I list the bar property (that was defined with defineProperty function?

Answer 1

Setting enumerable to true , will make it an enumerable property:

> Object.defineProperty(a, "b", { enumerable: true, get: function () { return "foo"; }})
{ foo: 'bar', b: [Getter] }
> for (k in a) { console.log(k); }
foo
b
> Object.keys(a)
[ 'foo', 'b' ]

enumerable

true if and only if this property shows up during enumeration of the properties on the corresponding object.

Defaults to false .

Answer 2

Both your question and your answer can be improved :

Your question is either :

• How can i define an enumerable property using Object.defineProperty ?

Answer : by setting, in the property parameters enumerable:true

OR

• How can i retrieve a property that was defined on an object as non enumerable ?

Answer : by using Object.getOwnPropertyNames , which will perform quite like keys , except... all own properties are returned, meaning : including the non-enumerable properties but not including prototype's properties (and obviously, no properties from the prototype chain).