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JavaScript JavaScript Foundations Functions Return Values

Carla Thomas
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Carla Thomas
Front End Web Development Techdegree Student 16,039 Points
Posted by Carla Thomas
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Carla Thomas
Front End Web Development Techdegree Student 16,039 Points

Return Values (Challenge)

I have been unsuccessfully working on this challenge for about 2 hours now. Please help.

Here are the instructions and what I have come up with.

Instructions

Around line 17, create a function named 'arrayCounter' that takes in a parameter which is an array. The function must return the length of an array passed in or 0 if a 'string', 'number' or 'undefined' value is passed in

This is how I have broken this task down

// Create a function named arrayCounter with parameter (argument) 'array'

function arrayCounter (array) {}

// Function must return the length of an array passed in or 0

return array.length;

My only guess is that the "typeof array" is incorrect. Here is what I am submitting

function arrayCounter (array) {

        if(typeof array === 'undefined'){
         array = 0;
        }

        return array.length;
      }

2 Answers

Jason Anello
Jason Anello
Courses Plus Student 94,610 Points

Hi Carla,

Your attempt is close and you have the right idea. You're checking if the value passed in is 'undefined' and that part is good except that you want to return 0 not set the array to zero.

Then you need 2 more if blocks just like that one to check if either a 'string' or a 'number' was passed in. So you'll want to have 3 if blocks and then all return 0.

At the end if you've made it past all those then you can return the length of the array.

Carla Thomas
seal-mask
.a{fill-rule:evenodd;}techdegree
Carla Thomas
Front End Web Development Techdegree Student 16,039 Points
Carla Thomas
.a{fill-rule:evenodd;}techdegree
Carla Thomas
Front End Web Development Techdegree Student 16,039 Points

This answer did not work for me. What am I doing wrong? Not fully understanding the "0" part in the instructions.

function arrayCounter (array) {

        if(typeof array === 'undefined',
           typeof array === 'string',
           typeof array === 'number'){
         return 0;
        }

        return array.length;
      }
Jason Anello
Jason Anello
Courses Plus Student 94,610 Points

Edit: Oops I did array = 0; instead of return 0

You need 3 separate if blocks.

if(typeof array === 'undefined'){
    return 0;
}
if(typeof array === 'string'){
    return 0;
}
if(typeof array === 'number'){
    return 0;
}
Jason Anello
Jason Anello
Courses Plus Student 94,610 Points

You're welcome. I forgot to answer "Not fully understanding the "0" part in the instructions"

The idea is that you want to return zero if any of those 3 types were passed in. This way you could test the return value of the function and if you see that it was zero then you know you must have passed in one of those 3 types, 'undefined', 'string', or 'number' and not an array.

Of course, you could pass in an empty array too and the function is also going to return zero.

Michelle Cannito
Michelle Cannito
8,992 Points
Michelle Cannito
Michelle Cannito
8,992 Points

You can combine the 3 if blocks into 1 by using || which means "OR"

Jason Anello
Jason Anello
Courses Plus Student 94,610 Points

This would be the better way to do it and would remove the repetition but I don't believe it was covered at this point in the course.

You can do this with the logical OR operator:

if(typeof array === 'undefined' || typeof array === 'string' || typeof array === 'number'){
    return 0;
}

This makes it more compact and removes the repetition.