Javascript !instanceof If Statement

Question

Enclose in parentheses and negate on the outside.

if(!(obj instanceof Array)) {
    //...
}

In this case, the order of precedence is important. See: Operator Precedence.

The ! operator precedes the instanceof operator.

if (!(obj instanceof Array)) {
    // do something
}

Is the correct way to check for this - as others have already answered. The other two tactics which have been suggested will not work and should be understood...

In the case of the ! operator without brackets.

if (!obj instanceof Array) {
    // do something
}

In this case, the order of precedence is important (https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Operators/Operator_Precedence). The ! operator precedes the instanceof operator. So, !obj evaluated to false first (it is equivalent to ! Boolean(obj)); then you are testing whether false instanceof Array, which is obviously negative.

In the case of the ! operator before the instanceof operator.

if (obj !instanceof Array) {
    // do something
}

This is a syntax error. Operators such as != are a single operator, as opposed to a NOT applied to an EQUALS. There is no such operator as !instanceof in the same way as there is no !< operator.

As explained in the other answers negation doesn't work because:

"the order of precedence is important"

But it is easy to forget the double parenthesis so you can make a habit of doing:

if(obj instanceof Array === false) {
    //The object is not an instance of Array
}

or

if(false === obj instanceof Array) {
    //The object is not an instance of Array
}

Try it here