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

Mark Cranny
Mark Cranny
3,019 Points
Posted by Mark Cranny
Mark Cranny
3,019 Points

Function to Variable

What is wrong with this code

function returnValue(anything) { return anything; }

var echo = returnValue();

returnValue("win");

2 Answers

0yzh 󠀠
0yzh 󠀠
17,276 Points
0yzh 󠀠
0yzh 󠀠
17,276 Points

Hey Mark,

The reason the code is not working is because you did not pass through an argument when assigning the function call returnValue(); to the variable echo. This makes the value of echo undefined. You can try something like this:

var echo = returnValue('win');
Mark Cranny
Mark Cranny
3,019 Points
Mark Cranny
Mark Cranny
3,019 Points

Thanks Huy Bui,

I got a little confused by the wording in the lesson.

Derek Gibson
Derek Gibson
6,531 Points
Derek Gibson
Derek Gibson
6,531 Points

What are you trying to accomplish with this code? Technically nothing is wrong with this code.

/* ORIGINAL CODE */
function returnValue(anything) { 
  return anything; 
}

var echo = returnValue();

returnValue("win");

Your code first creates a function named returnValue with a parameter named "anything" IF your function is called, the code inside the function will run, that code "return anything" will return a the value of whatever argument you pass to it.

Second, you created a variable named echo which is equaled to your returnValue function with NO argument passed to it.

Third, you call the function correctly passing "win" as a argument to it, which executes the code within the function, returning "win". If you changed the last line of your code to:

console.log( returnValue("win") );

and opened the console, you would see the string 'win' in the console.

I am still learning myself so I hope this answer helps and is factual. I don't understand the purpose of creating the echo variable in this code but like I previously mentioned I am also learning. Good luck!

Mark Cranny
Mark Cranny
3,019 Points
Mark Cranny
Mark Cranny
3,019 Points

Thanks, one of the tasks in the Javascript wasn't completing and throwing an error. Huy Bui was correct I had to assign the value as an argument at the time of assigning it to the variable like:

var echo = returnValue('win');

Thanks for the input