Is the string uppercase? guys i can’t solve this challenge for some reason that i don’t know. Any form of help would be appreciated. my code:

String.prototype.isUpperCase = function() {

if (str === str.toUpperCase()) { 
  return true;
else {
  return false;

let str = "hello I AM DONALD";

You are seeing an error message like:

ReferenceError: str is not defined

Why? Because the challenge is not defining str anywhere and you attempt to reference str without it being defined. You need to think about how to reference the the string upon the isUpperCase method is called.


Just like Randell says above, you need to use a keyword inside of the function that the prototype would have access to.

String.prototype.isUpperCase = function() {
if (this == this.toUpperCase()){
return true} else return false;

so use “this” only or "JSON.stringify(this) ?

not sure what prototype refers to

In JavaScript there is such a thing called a prototype chain, which is essentially a way of creating objects where all of their children share the same methods.

In this example you are modifying the prototype of the String class in JavaScript. By doing so, you are giving this new method to each and every string that is created within this scope of your document.

Using only this does not work because this is a String object which sort of acts just like it is in reality, being an Array of characters. By using JSON.stringify you are changing the value of this to an actual string, thus allowing you to compare it to a version with the method toUpperCase() applied to it.

So the str variable that i assigned is a prototype of String after i did this “str.isUpperCase()” ?

When i console log “this”. it shows --> [String: ‘C’]

so basically the characters are in an array as you said and i tried this[0] but it only gives first character. i want the whole string so i guess JSON.stringify works since it turns the array into a whole string of the characters inside it.

thanks buddy it worked.

valueOf or toString is your friend here.

1 Like

You are correct to ensure its not the object wrapper, but
don’t use JSON.stringify unless it’s actually necessary, toString is fine. Or == instead of ===, which will perform the coercion automatically but it’s less obvious what is happening:

if (this == this.toUppercase()) {


if (this.toString() === this.toUppercase()) {

Also, strings are not arrays of characters (thank god!), JS is not C


Good point. I forgot about the use of == in this case.

wait… if (this == this.toUppercase())

does == convert the array format to string for us?

edit: cuz from what i understand it does type conversion, so we can compare numbers with strings. but this?

Sort of, it will do some conversion/coercion. You need to understand the rules surrounding it though: as a rule never use ==, always use === because if you don’t understand the difference it will cause errors in your code.

Also it’s not an array. For a fractionally small amount of time, strings get wrapped in objects when string methods are used (as in this case), you almost never ever have to think about it.

Thanks for the clarification!

I dunno why my testing pushed me into using the JSON.stringify method. For some reason I wasn’t able to get true without it. I updated the code to reflect the change though.

I was up a bit late though maybe I was just super tired lol!

hey if you don’t mind, can you help me out on this one? i can’t tell the difference… between both codes

thanks i didn’t know that. can you help me out on this algorithm? A String of Sorts

i get that strings are wrapped in objects when string methods are used. but if it isn’t an array then what is it? it shows as an array when no methods in used.

What specific thing are you talking about here? Because a string will be a string if no methods are used, not an array, only an array will show as an array.

when i console log “this”. it shows as an array. it shows ----> [String: “C”]
edit: also would u be so kind to help me out on “a string of sorts” that i just posted? i really need help on that… maybe u can tell the difference because i can’t

It’s not showing as an array, that’s what the console uses to show the type/value of an on object. The type is String, the value is c.

okay… welp im assuming u dw to help me out on the algorithm

:+1: replied in that thread.