Javascript prototype inheritance doesn't call expected method

Question

Why does this print out 'bye' instead of 'hello'? According to the inheritance chain as is described in this blog post, I would've thought it would log 'hello'.

http://sporto.github.io/blog/2013/02/22/a-plain-english-guide-to-javascript-prototypes/

class Test {
  hello() {
    console.log('hello')
  }
}

Test.prototype.hello = function(){
  console.log('bye')
}

const t = new Test
t.hello()

Why do you overwrite Test.prototype.hello if you don't want that? What else did you expect to happen with the bye-function? — Bergi
– Bergi, Commented May 17, 2017 at 1:35
Do you know how class syntax works? What do you think does the inheritance chain of t look like according to the article you read? — Bergi
– Bergi, Commented May 17, 2017 at 1:37

theprogrammer · Accepted Answer · 2017-05-17 00:39:06Z

3

You are rewriting the definition of hello on the "prototype". When you do class Test () ... hello is the equivalent of

Test.prototype.hello

The class syntax is mostly sugar on top of the normal prototypical definition of a function.

answered May 17, 2017 at 0:39

theprogrammer

2,7341 gold badge19 silver badges13 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Javascript prototype inheritance doesn't call expected method

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related