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

javascript problem

I don't understand this part

script.js
function returnValue(accept) {
 return accept;
  var echo = 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>

1 Answer

Collin Halliday
seal-mask
.a{fill-rule:evenodd;}techdegree
Collin Halliday
Full Stack JavaScript Techdegree Student 17,491 Points

Hey, Yisroel.

You want to make sure that your variable declaration and your call to the returnValue() function are outside of the returnValue() function. The way you have it written currently, nothing will happen because your function is never called. The returnValue() function will not execute until called from somewhere outside of itself. However, even if you called the function as you have it written currently, your variable declaration for "echo" and its initialization to "returnValue('hello')" will never run because that line of code is placed after the function's return statement. After that return statement, your program will exit the function and any code included afterword will be ignored.

Try moving the following line outside of and after the function: "var echo = returnValue('hello');"

I hope that is helpful. Best of luck!

Yes it did, thank you so much!