isPrototypeOf和__proto__的结果不同

Question

The following two expressions:

"abc".__proto__.__proto__ === Object.prototype  // true
Object.prototype.isPrototypeOf("abc")           // false

The first expression proves that Object.prototype lies in the prototype chain of "abc". However, the second expression gets an opposite result.

I am very confused. Hope anyone can explain.

Answer 1

"abc" is not an object. When you evaluate "abc".__proto__ , a String wrapper object is implicitly constructed to retrieve the prototype of, and Object.prototype is in that wrapper object's prototype chain.

Object.prototype.isPrototypeOf("abc") doesn't construct a wrapper object. It just looks at "abc" , sees that "abc" isn't an object and has no prototype chain, and returns false. You can see this in the ECMAScript spec (version 6):

When the isPrototypeOf method is called with argument V, the following steps are taken:

1. If Type(V) is not Object, return false.