Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

JavaScript JavaScript Basics (Retired) Creating Reusable Code with Functions Passing an Argument to a Function

Erin Van Ess
Erin Van Ess
8,023 Points
Posted by Erin Van Ess
Erin Van Ess
8,023 Points

Cant figure this out - I think because of the way the function is named returnValue.

I know I am missing something very simple... its like a trick question with the naming.

script.js 
function returnValue(anything){
  var echo = returnValue(anything);

}
return echo ('silly');
index.html 
<!DOCTYPE HTML>
<html>
<head>
  <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
  <title>JavaScript Basics</title>
</head>
<body>
<script src="script.js"></script>
</body>
</html>

1 Answer

Alexander Davison
Alexander Davison
65,469 Points 
function returnValue(anything){
   return anything;
}

All you need to do is define a function that takes one parameter and then returns that parameter. :)

~Alex

Erin Van Ess
Erin Van Ess
8,023 Points
Erin Van Ess
Erin Van Ess
8,023 Points

I did get the first part, but thanks so much for your answer. It's the second step that is confusing me. Here are the instructions for that - After your newly created returnValue function, create a new variable named echo. Set the value of echo to be the results from calling the returnValue function. When you call the returnValue function, make sure to pass in any string you'd like for the parameter.

Nevermind - I finally figured it out. For some reason I was putting the var echo inside the function instead of after the function. Jeez... so easy!! It kills me when the thing I'm missing is so simple!!!

Alexander Davison
Alexander Davison
65,469 Points

Heh :)