Javascript new array prototype in new object constructor

Question

If I do:

Array.prototype.test = "test"
Array.prototype.t = function() {return "hello"}

Every new Array will have the property test and the method t.

How can I do the same without affecting all Arrays?

Like:

Names = function(arr){
  // Contacts constructor must be the same of Array 
  // but add the property test and the function t
}
z=new Names(["john","andrew"])

So that z.test will return "test" and z.t() will return "hello"? (but Array.test and Array.t would stay undefined)

I explain better:

Array.prototype.t="test";
Array.prototype.test = function(){ return "hello";}

z=new Array("john", "andrew")
console.log(z);

But this affects ALL arrays. I want the same but with a new constructor Names that inherits Array constructor.

"// Contacts constructor must be the same of Array " this part is not clear. Do you mean Names instances should inherit Array.prototype as well? — Yury Tarabanko
– Yury Tarabanko, Commented Jul 2, 2018 at 7:09
I want a prototype that does not affect all arrays but only itself.. Array.prototype.test = "test" Array.prototype.t = function() {return "hello"} affects all arrays I wish to do Names.prototype.test = "test" Names.prototype.t = function() {return "hello"} where Names inherits everything else from Array prototype — Zibri
– Zibri, Commented Jul 2, 2018 at 8:44

Isaac · Accepted Answer · 2018-07-02 09:53:28Z

2

class Names extends Array {
  constructor(...args) {
    super(...args);
  }
}

Names.prototype.t = 'test';

let z = new Names("john", "andrew")
z.push('Amanda')

console.log(z.t)
console.log(z)

You can easily set it at Names.prototype

edited Jul 2, 2018 at 9:53

answered Jul 2, 2018 at 6:25

Isaac

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5 Comments

Zibri Over a year ago

not what I intended to do. I want the same effect of the Array.prototype example but affecting only Names and not all arrays.

Isaac Over a year ago

@Zibri: Please have a look at my updated solution. z can access the push method just like Array and z can access t as you wanted. And it doesn't pollute Array, as you wanted

Zibri Over a year ago

perfect! I was getting crazy!

Zibri Over a year ago

the instructions I was searching for where the first lines class Names extends Array { constructor(...args) { super(...args); } } THANKS!

Isaac Over a year ago

@Zibri: Glad to help! I know that feeling :D

ZER0 · Accepted Answer · 2018-07-02 08:43:28Z

1

Can't you just extend Array?

class Names extends Array {
  constructor(...args) {
    super(...args);
    this.t = "test";
  }

  test() { return "hello" }
}

let z = new Names("john", "andrew")

edited Jul 2, 2018 at 8:43

answered Jul 2, 2018 at 8:28

ZER0

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2 Comments

Zibri Over a year ago

Nope.. executing your example returns: Names [t: "test"] instead of ["john","andrew"]

ZER0 Over a year ago

@Zibri Sorry, my bad, I actually forgot to pass the arguments to super

Ivan · Accepted Answer · 2018-07-02 08:47:43Z

1

Here is a crude implementation:

function Names(arr) {
  this.contacts = enhanceArray(arr);
}

function enhanceArray(arr) {
  arr.test = 'helloProp';
  arr.t = function() {
    return 'helloFunc'
  }
  return arr;
}
let z = new Names(["john", "andrew"]);

console.log(z.contacts[0]);
console.log(z.contacts.test);
console.log(z.contacts.t());

edited Jul 2, 2018 at 8:47

Ivan

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answered Jul 2, 2018 at 8:45

Rahil Ahmad

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Comments

KooiInc · Accepted Answer · 2018-07-02 09:08:29Z

0

You can create your own extended Array constructor factory, something like

(() => {
  const myArr = XArray();
  let a = myArr(["John", "Mary", "Michael"]);
  console.log(`myArr(["John", "Mary", "Michael"]).sayHi(1): ${a.sayHi(1)}`);
  console.log("demo: myArr(1, 2, 3) throws an error");
  let b = myArr(1, 2, 3); // throws
  
  // an extended array
  function XArray() {
    const Arr = function(arr) {
      if (arr.constructor !== Array) {
        throw new TypeError("Expected an array");
      }
      this.arr = arr;
    };
    Arr.prototype = {
      sayHi: function (i) { return `hi ${this.arr[i]}`; }
    };
    return arr => new Arr(arr);
  }
})();

edited Jul 2, 2018 at 9:08

answered Jul 2, 2018 at 7:10

KooiInc

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4 Comments

Ivan Over a year ago

could you check your code, it seems there is a problem when executing. ;)

KooiInc Over a year ago

@Ivan that's by design ;P

Ivan Over a year ago

oh sorry, I didn't pay enough attention!

KooiInc Over a year ago

No worries, the answer was evidently not clear enough. Edited it a bit.

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