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Yisroel stadler
2,910 Pointsjavascript problem
I don't understand this part
function returnValue(accept) {
return accept;
var echo = returnValue('hello');
}
<!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
Full Stack JavaScript Techdegree Student 17,491 PointsHey, 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!
Yisroel stadler
2,910 PointsYisroel stadler
2,910 PointsYes it did, thank you so much!