Is not a function using prototype in js vanilla

Question

This is my error, but I have some functions like that and they work. What am I doing wrong?

I try to change the prototype's name but it's the same error

function Pendejo(nickname, edad, pendejo) {
  this.nickname = nickname;
  this.edad = edad;
  this.pendejo = pendejo;
}

Pendejo.prototype.show = function() {
  return `Soy ${this.nickname} y soy ${this.pendejo}% pendejo`;
}

let ryu = new Pendejo('Ryu', 22, 100);
console.log(Pendejo.show());

`ryu.show()`, not `Pendejo.show()`. Voting to close as a typo — Rory McCrossan, Feb 25 '23 at 15:56
Does this answer your question? [Difference between adding function to prototype and object literal in javascript](https://stackoverflow.com/questions/20724533/difference-between-adding-function-to-prototype-and-object-literal-in-javascript) — Wyck, Feb 25 '23 at 15:56

score 1 · Accepted Answer · answered Feb 25 '23 at 15:55

You need to invoke the function from the object ryu

function Pendejo(nickname, edad, pendejo) {
  this.nickname = nickname;
  this.edad = edad;
  this.pendejo = pendejo;
}

Pendejo.prototype.show = function() {
  return `Soy ${this.nickname} y soy ${this.pendejo}% pendejo`;
}

let ryu = new Pendejo('Ryu', 22, 100);
console.log(ryu.show());

score 1 · Answer 2 · answered Feb 25 '23 at 15:56

You should call show() on the object, not the function:

function Pendejo(nickname, edad, pendejo) {
  this.nickname = nickname;
  this.edad = edad;
  this.pendejo = pendejo;
}

Pendejo.prototype.show = function() {
  return `Soy ${this.nickname} y soy ${this.pendejo}% pendejo`;
}

let ryu = new Pendejo('Ryu', 22, 100);
console.log(ryu.show());

Is not a function using prototype in js vanilla

2 Answers2