How to check if x is an Object but not a String Object

Question

What I have so far.

const isNotNullObject = function (x) {
    return (typeof x === "object" && x !== null);
};

It works fine for arrays and objects. But for String objects too !

isNotNullObject(String(5))
false
isNotNullObject(new String(5))
true

What I want is false for any type of string. Note that I don't have control of the calling code. I can't remove new myself. I need a solution that does not create a new String just to check for equality if possible for performance reasons.

But string objects *are* objects. Treat them as such and assume that your caller knows not to wrap strings. — Bergi, Sep 05 '16 at 16:26
@ifvictr: If `typeof x == "object"` is `true`, then we already know `typeof x != "string"` is also `true`. The value of `typeof x` isn't going to change after the first comparison. — , Sep 05 '16 at 17:45

score 4 · Accepted Answer · answered Sep 05 '16 at 16:23

4

Use instance of

return (typeof x === "object" && !(x instanceof String) && x !== null)

const isNotNullObject = function(x) {
  return (typeof x === "object" && !(x instanceof String) && x !== null);
};

console.log(
  isNotNullObject(String(5)),
  isNotNullObject(new String(5))
)

answered Sep 05 '16 at 16:23

Pranav C Balan

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@Bergi : what you mean by frames – Pranav C Balan Sep 05 '16 at 16:29
See http://perfectionkills.com/instanceof-considered-harmful-or-how-to-write-a-robust-isarray/ or http://stackoverflow.com/q/22289727/1048572 – Bergi Sep 05 '16 at 16:31

score 0 · Answer 2 · answered Sep 05 '16 at 16:34

There are many ways to check type of Object/String/Array.

using type of X operator using
Object.prototype.toString.apply(X)

//Its performance is worst
Object.getPrototypeOf(X)
X.constructor.

Where X can be Object/Array/String/Number or Anything. For performance Comparison Please see below image

How to check if x is an Object but not a String Object

2 Answers2