Passing an array as a parameter to a function to pick a random string

Question

I am currently doing a javascript course on udemy and ran into a problem with a small coding challenge. The code is supposed to pick a random person from an array. My code works on their test platform, but i cant get it to work in the chrome developer tools.

var names = ["Angela", "Ben", "Jenny", "Michael", "Chloe"];
 
function whosPaying(names) {
    var randomNumber = Math.random();
    var arrayLength = names.length;
    var hasToPay = Math.floor((arrayLength * randomNumber));
 
    return names[hasToPay] + " is going to buy lunch today!"
}
 
whosPaying();

If i write my code like this, it works in the test environment when I submit the code without the array "names", but in chrome I get the following error:

index.js:5 Uncaught TypeError: Cannot read properties of undefined (reading 'length')
    at whosPaying (VM165 index.js:5)
    at VM165 index.js:11

If i define the array inside the function, it works, but that is not really what I'm supposed to do.

Im wondering how i can make the function use the array that is defined outside of the function. In the exercise instructions the array and the way it is passed on to the function are predefined exactly the same way i did it.

I could not find a solution to this that does not include completely different syntax or stuff I am not yet supposed to know at this point in the course.

Just send the argument whosPaying(names);

ABDULLOKH MUKHAMMADJONOV
– ABDULLOKH MUKHAMMADJONOV

2022-01-04 20:21:45 +00:00
Commented Jan 4, 2022 at 20:21 — ABDULLOKH MUKHAMMADJONOV
– ABDULLOKH MUKHAMMADJONOV, Commented Jan 4, 2022 at 20:21

Micael Levi · Accepted Answer · 2022-01-04 20:20:46Z

4

you've called whosPaying without any args, then names end up being undefined.

Do this instead: whosPaying(names), or drop the names parameter of whosPaying

answered Jan 4, 2022 at 20:20

Micael Levi

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2 Comments

Dave Newton Over a year ago

@TheHansDampf FWIW the easiest thing to do is put a console.log (or breakpoint) immediately inside the function to check your assumptions around the the parameter(s). Once you saw that it was undefined the logical place to look at where it's being called.

Arman Ebrahimi Over a year ago

In this case, isn't this better to be written without argument & parameter? Or if be both is better?

Maik Lowrey · Accepted Answer · 2022-01-04 20:22:11Z

2

You overwrite global names if you pass a parameter with the same name.

var names = ["Angela", "Ben", "Jenny", "Michael", "Chloe"];
 
function whosPaying() {    
    var randomNumber = Math.random();
    var arrayLength = names.length;
    var hasToPay = Math.floor((arrayLength * randomNumber));
 
    return names[hasToPay] + " is going to buy lunch today!"
}
 
let r = whosPaying();
console.log(r);

answered Jan 4, 2022 at 20:22

Maik Lowrey

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Comments

Sercan · Accepted Answer · 2022-01-04 20:23:55Z

2

You can pass parameters to the function and print the return value:

function whosPaying(names)
{
    var randomNumber = Math.random();
    var arrayLength = names.length;
    var hasToPay = Math.floor((arrayLength * randomNumber)); 
    return names[hasToPay] + " is going to buy lunch today!"
}

var names = ["Angela", "Ben", "Jenny", "Michael", "Chloe"];

console.log(whosPaying(names))

answered Jan 4, 2022 at 20:23

Sercan

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Comments

Relcode · Accepted Answer · 2022-01-09 19:42:05Z

As it has been pointed out that you didn't provide the names argument when you invoked your whosPaying() function... because you used var to declare the names variables, you can do this without passing any arguemnt because of scoping.

var names = ["Angela", "Ben", "Jenny", "Michael", "Chloe"];
 
function whosPaying() {
    var randomNumber = Math.random();
    var arrayLength = names.length;
    var hasToPay = Math.floor((arrayLength * randomNumber));
 
    console.log(names[hasToPay] + " is going to buy lunch today!");
}
 
whosPaying();

Also this...

var randomNumber = Math.random();
var arrayLength = names.length;
var hasToPay = Math.floor((arrayLength * randomNumber));

Can be converted to this:

var hasToPay = Math.floor(Math.random() * names.length);

Which is a single line instead of three.

user17597013 · Accepted Answer · 2022-01-04 20:22:04Z

0

I assume that you want to pick a random index from your array , you can try this

function generateRandom(Array) {
  let random = Math.floor(Math.random() * Array.length)
  return Array[random]
}

answered Jan 4, 2022 at 20:22

user17597013

357 bronze badges

1 Comment

Dave Newton Over a year ago

Please don't, however, name a parameter after an existing thing (and better to stick to standard JS naming conventions).

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