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

Posted by Samuel Day
Samuel Day
772 Points

Im about to flip a lid. Your instructions are AGAIN unclear and there is no help to clarify what I am to do.

Its telling me I need to store the value of the function "returnValue" in the variable echo. Which im clearly doing. I aslo tried: var echo = argument; var echo = returnValue(argument);

script.js 
function returnValue(argument){
  return argument;
}
var echo = returnValue();
returnValue('hello');
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>

2 Answers

Krishna Pratap Chouhan
Krishna Pratap Chouhan
15,203 Points

You have defined the function correctly.

Now function takes one argument and it returns it. so whenever we call the function we need to pass an argument into it.

Also we have to get the returned value stored into the echo variable.

here is how to do it.

function returnValue(arg){
  return arg;
}

var echo = returnValue("something");
Samuel Day
Samuel Day
772 Points

Bless you and your knowledge.

Jon Mirow
Jon Mirow
9,864 Points
Jon Mirow
Jon Mirow
9,864 Points

Hi there. Sorry to hear about the frustration. Hope I can help by breaking down the question:

"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.

The function will accept the string you pass to it and return it back (using the return statement). Then your echo variable will be set with what is returned, the results."

So we have two steps and a reminder. Step 1: create a new vaiable named echo - check Step 2: call the returnValue function and asign the value it returns to echo - check Reminder: Don't forget returnValue() expects an input - that's where the confusion comes in.

In your code, you assign echo to the output of return value, but without passing any arguments into returnValue, echo ends up as 'undefined'. You then call returnValue with a parameter but don't store the output so it is discarded.