CODEWITHSUNDEEP
javascriptprototype
zurfyx
zurfyx

Reputation: 32767

Object.prototype.isPrototypeOf vs. obj.prototype.isPrototypeOf

I am not looking forward the differences between one and the other. Airbnb already does a great job explaining about it on the style guides repository.

Considering a trivial class and its implementation like the following:

class C1 {}
const c1Imp = new C1();

Where .prototype should be inherited from Object.

Why isn't the following equivalent?

console.info(Object.prototype.isPrototypeOf.call(C1, c1Imp)); // false
console.info(C1.prototype.isPrototypeOf(c1Imp)); // true


 (() => {
	class C1 {}
	const c1Imp = new C1();

	console.info(c1Imp.constructor.name);
	console.info(`${Object.prototype.isPrototypeOf.call(C1, c1Imp)} (should be true)`);
	console.info(`${C1.prototype.isPrototypeOf(c1Imp)} (should be true)`);
	 
	class C2 {}
	const c2Imp = new C2();
	 
	console.info(c2Imp.constructor.name);
	console.info(`${Object.prototype.isPrototypeOf.call(C1, c2Imp)} (should be false)`);
	console.info(`${C1.prototype.isPrototypeOf(c2Imp)} (should be false)`);
})();

PS: the question title isn't very clear, feel free to edit as appropriate.

Upvotes: 1

Views: 115

Answers (1)

Josh Beam
Josh Beam

Reputation: 19772

You should do this instead:

Object.prototype.isPrototypeOf.call(C1.prototype, c1Imp) // true

In your first example, you were calling the Object.prototype.isPrototypeOf method on C1 itself, while in the second example, you were calling isProtoTypeOf on the C1.prototype. It's just some tricky semantics.

The fix as I showed above is to called Object.prototype.isPrototypeOf on the C1.prototype itself.

Check out the updated snippet:

 (() => {
	class C1 {}
	const c1Imp = new C1();

	console.info(c1Imp.constructor.name);
	console.info(`${Object.prototype.isPrototypeOf.call(C1.prototype, c1Imp)} (should be true)`);
	console.info(`${C1.prototype.isPrototypeOf(c1Imp)} (should be true)`);
	 
	class C2 {}
	const c2Imp = new C2();
	 
	console.info(c2Imp.constructor.name);
	console.info(`${Object.prototype.isPrototypeOf.call(C1, c2Imp)} (should be false)`);
	console.info(`${C1.prototype.isPrototypeOf(c2Imp)} (should be false)`);
})();

Upvotes: 2

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