Best way to “negate” an instanceof

Question

I was thinking if there exists a better/nicer way to negate an instanceof in Java. Actually, I\'m doing something like:

if(!(str instanceof String))         


        

Answer 1

You can achieve by doing below way.. just add a condition by adding bracket if(!(condition with instanceOf)) with the whole condition by adding ! operator at the start just the way mentioned in below code snippets.

if(!(str instanceof String)) { /* do Something */ } // COMPILATION WORK

instead of

if(str !instanceof String) { /* do Something */ } // COMPILATION FAIL

Answer 2

No, there is no better way; yours is canonical.

Answer 3

I agree that in most cases the if (!(x instanceof Y)) {...} is the best approach, but in some cases creating an isY(x) function so you can if (!isY(x)) {...} is worthwhile.

I'm a typescript novice, and I've bumped into this S/O question a bunch of times over the last few weeks, so for the googlers the typescript way to do this is to create a typeguard like this:

typeGuards.ts

export function isHTMLInputElement (value: any): value is HTMLInputElement {
  return value instanceof HTMLInputElement
}

usage

if (!isHTMLInputElement(x)) throw new RangeError()
// do something with an HTMLInputElement

I guess the only reason why this might be appropriate in typescript and not regular js is that typeguards are a common convention, so if you're writing them for other interfaces, it's reasonable / understandable / natural to write them for classes too.

There's more detail about user defined type guards like this in the docs