JavaScript object will not print

Question

I am trying to print out my object but it doesn't seem to be working. I'm very new to coding and am not sure as to what I am doing wrong here. Please help!

function Coffee(name, roast, location) {
    this.name = name;
    this.roast = roast;
    this.location = location;

    var esspresso = new Coffee("Starbuck's Esspresso Blend", "Dark", "Seattle");
    var cappuccino = new Coffee(coffee.name(espresso), "Dark", "Seattle");

    var CoffeeType = function() {
        return coffee.name + " " + coffee.roast + " from " + coffee.location;
    };
}
document.write(CoffeeType(cappuccino));

`CoffeeType` doesn't exist where you are trying to access it. Did you mean to put the `}` after `this.location = location;` instead of before `document.write`? It doesn't make a lot of sense to call `Coffee` instead itself. `CoffeeType` also doesn't expect any argument and there is no variable with name `coffee`. Overall this seems to be arbitrarily stitched together. Which tutorial are you following? I suggest to read http://eloquentjavascript.net/ instead. — Felix Kling, Nov 06 '15 at 19:12
Have some literature on scope! http://toddmotto.com/everything-you-wanted-to-know-about-javascript-scope/ — Xogle, Nov 06 '15 at 19:15
I'm not using a tutorial... Just trying to figure things out on my own... Thanks for the insight! — Dustin Baker, Nov 06 '15 at 19:31
_"it doesn't seem to be working"_ is not a good enough problem statement. — Lightness Races in Orbit, Nov 06 '15 at 21:37

score 3 · Accepted Answer · edited May 23 '17 at 12:15

If you're shooting for an object oriented approach, you should be defining your functions on the prototype chain.

function Coffee(name, roast, location) {
    this.name = name;
    this.roast = roast;
    this.location = location;
}

Coffee.prototype.getCoffeeType = function() {
    return this.name + " " + this.roast + " from " + this.location;
};

var esspresso = new Coffee("Starbuck's Esspresso Blend", "Dark", "Seattle");
var cappuccino = new Coffee(esspresso.name, "Dark", "Seattle");

document.write(cappuccino.getCoffeeType());

The code above will have the following effects:

function Coffee(...) is the class's main constructor.
Coffee.prototype.getCoffeeType is a function that each instance of the Coffee class will have. This function will be able to access the this variables of the class (the class members).
esspresso and cappuccino are instances of the Coffee class, instantiated somewhere in program code (outside of the class definition / constructor / prototype functions).
You can then call functions defined on the class's prototype chain on these instance objects.
The output of the above code will give you "Starbuck's Esspresso Blend Dark from Seattle"

Note that the additional benefit of defining a function on the prototype is that all instances will share the same getCoffeeType function instead of having a separate function (which does the same thing) on each instance. (see Use of 'prototype' vs. 'this' in JavaScript?). If you're running a large coffee chain and producing millions of coffees, then you'll probably save a fair amount of resources (i.e. memory).

Wow! Thanks a ton for the deep explanation. You don't know how much I appreciate it! — Dustin Baker, Nov 06 '15 at 19:32

score 1 · Answer 2 · answered Nov 06 '15 at 19:14

CoffeeType is defined only in the scope of Coffee, and can't be used outside of the { }. You also need to pass in the coffee as an argument to use it. var CoffeeType = function( coffee ) {.

Change your program to define CoffeeType in a higher scope so it can be used outside of Coffee's body:

function CoffeeType(coffee) {
    return coffee.name + " " + coffee.roast + " from " + coffee.location;
};

function Coffee(name, roast, location) {
    ...

score 1 · Answer 3 · edited Nov 06 '15 at 21:38

Try this:

function Coffee(name, roast, location) {

    this.name = name;
    this.roast = roast;
    this.location = location;
    this.CoffeeType = function() {
        return this.name + " " + this.roast + " from " + this.location;
    };

}

var esspresso = new Coffee("Starbuck's Esspresso Blend", "Dark", "Seattle");
var cappuccino = new Coffee("espresso", "Dark", "Seattle");

console.log(esspresso.CoffeeType())

JavaScript object will not print

3 Answers3