how to properly call functions in prototype in js

Question

How do I properly call the functions inside pretest?

I get this error: Uncaught TypeError: b.testmenow is not a function

    var pretest = function () {
        var MAX_NUM = 250.0;
    
        var prebase = function (NEW_NUM) {
            this.NEW_NUM = NEW_NUM ? NEW_NUM : true;
        };
    
        prebase.prototype.testmenow = function () {
            return this.NEW_NUM;
        };
        
        return prebase;
    };
    
    var b = new pretest(111);
    console.log(b.testmenow());

Part of me says you need to return new prebase() but it's super strange how you're doing this. Try searching to learn more on prototypes and constructors. — Chris
– Chris, Commented Jul 3, 2018 at 0:56
pretest is a function that returns a constructor, it's not a constructor itself. Not sure why you are doing this, but with minimal changes to your code: var b = new (pretest())(111);. — Felix Kling
– Felix Kling, Commented Jul 3, 2018 at 1:00
Curious to know why you don't just use prebase directly — what is pretest's contribution? — Mark
– Mark, Commented Jul 3, 2018 at 1:04

Chris · Accepted Answer · 2018-07-03 01:05:54Z

1

You need to accept your input into new pretest(111) by adding n. And then you must instantiate your prebase constructor using n.

    var pretest = function (n) {
        var MAX_NUM = 250.0;
    
        var prebase = function (NEW_NUM) {
            this.NEW_NUM = NEW_NUM ? NEW_NUM : true;
        };
    
        prebase.prototype.testmenow = function () {
            return this.NEW_NUM;
        };
        
        return new prebase(n);
    };
    
    var b = pretest(111);
    console.log(b.testmenow());

It is strange that you have two constructors here, you can surely do this with one.

As Felix has deftly mentioned, you can call pretest(111) instead of new pretest(111).

edited Jul 3, 2018 at 1:05

answered Jul 3, 2018 at 1:03

Chris

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2 Comments

Felix Kling Over a year ago

There is no reason to call pretest with new. It's rather confusing. Just do var b = pretest(111);.

Boy Over a year ago

so If I have multiple parameters I need to return new prebase(n,x,y,z); ?

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