CODEWITHSUNDEEP
javascript
zer0uno
zer0uno

Reputation: 8030

Why do Object.prototype and Object.getOwnPropertyNames(Object.prototype) return different things?

Object.prototype
=> {}
   Object.getOwnPropertyNames(Object.prototype)
=> [ 'constructor',
  'toString',
  'toLocaleString',
  'valueOf',
  'hasOwnProperty',
  'isPrototypeOf',
  'propertyIsEnumerable',
  '__defineGetter__',
  '__lookupGetter__',
  '__defineSetter__',
  '__lookupSetter__',
  '__proto__' ]
   var x = {foo:2}
   x
=> { foo: 2 }
   Object.getOwnPropertyNames(x)
=> [ 'foo' ]

Why does Object.prototype is empty but calling Object.getOwnPropertyNames(Object.prototype) it is really not?

Upvotes: 1

Views: 76

Answers (1)

T.J. Crowder
T.J. Crowder

Reputation: 1074989

It's probably down to the fact that all of Object.prototype's properties are non-enumerable. It'll vary by console, but apparently the console you're using isn't showing non-enumerable properties when you ask it to show you Object.prototype. In contrast, getOwnPropertyNames returns an array containing the names of all "own" properties of the object you pass in whose names are strings, regardless of whether they're enumerable; your console is then showing you the contents of that array. (If you used Object.keys(Object.prototype) instead, you'd get an empty array, because Object.keys only gives you the names of enumerable properties.)

In your x example, the object has an enumerable foo property. I suspect if you did this:

var x = {};
Object.defineProperty(x, "foo", {value: 2});
x

...in your console, you'd see {} like you do for Object.prototype, because that would define foo as a non-enumerable property.

Upvotes: 2

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