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.
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)`); })();
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.