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

Posted by Chikanma Ibeh
Chikanma Ibeh
1,396 Points

Why is this wrong?

unction returnValue (price) { var echo = return price; }

return('cheap');

script.js 
function returnValue (price) {
  var echo = return price;
}

return('cheap');
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>

3 Answers

Gav Cooper
Gav Cooper
4,261 Points

There's no need, as far as I'm aware, to set the return value of a function to its own variable, you can just return price straight away.

Also, to call the function you need to use its name e,g, returnValue("cheap")

Brandyn Lordi
Brandyn Lordi
17,778 Points
Brandyn Lordi
Brandyn Lordi
17,778 Points

your script is not written correctly. Check out my example below.

function returnValue(price) {
  return price;
}
returnValue('value');

Notice return price is nested within my function in my example. After creating the function, you call the function, and pass the argument (or parameter) - in this case, i pass value. It returns my given parameter because i tell it to do so in my function with return price

You could alter your current function to work by making the following adjustments:

function returnValue (price) {
  var echo = price;
  return echo;
}
returnValue('cheap');

Does this make sense?

Zack Jackson
Zack Jackson
30,220 Points
Zack Jackson
Zack Jackson
30,220 Points

Not sure why Brandyn was down voted. His response was correct.

This part of your code sets up a function called returnValue where you are passing into the function an argument named price.

function returnValue (price) {
  ...
}

This part of your code is just incorrect syntax. Not sure what you're trying to do here, but you passed in price and you want to return it and assign it to a variable called echo? Why? Maybe you can explain what you're trying to do with this code so that we can better help you get it working.

  var echo = return price;

This part of your code is also just incorrect syntax. Your function is called "returnValue", not "return". The cheap that you're passing in is the price arg from above. Your function should do something with this. See example below. 

return('cheap');

For instance, if you had the function return a sentence or whatever that included the price that was passed in. This passed in arg is being concatenated onto this string. The result would be "This item is cheap." being returned.

function returnValue(price) {
  return('This item is ' + price + '.');
}

returnValue('cheap');