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
– Bergi, Commented Sep 5, 2016 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. — user1106925
– user1106925, Commented Sep 5, 2016 at 17:45

Pranav C Balan · Accepted Answer · 2016-09-05 16:23:49Z

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 5, 2016 at 16:23

Pranav C Balan

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2 Comments

Pranav C Balan Over a year ago

@Bergi : what you mean by frames

Bergi Over a year ago

See perfectionkills.com/… or stackoverflow.com/q/22289727/1048572

Vishnu Shekhawat · Accepted Answer · 2016-09-05 16:34:18Z

0

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

answered Sep 5, 2016 at 16:34

Vishnu Shekhawat

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