javascript - Let a function "return" the super function? - Stack Overflow

Given is the following code:

function two() {
    return "success";
}

function one() {
    two();
    return "fail";
}

If you test the code by calling function one(), you will always get "fail".

The question is, how can I return "success" in function one() by only calling function two()?

Is that even possible?

Regards

Given is the following code:

function two() {
    return "success";
}

function one() {
    two();
    return "fail";
}

If you test the code by calling function one(), you will always get "fail".

The question is, how can I return "success" in function one() by only calling function two()?

Is that even possible?

Regards

• javascript
• function
• return
• super

Share Improve this question Follow asked Oct 13, 2010 at 13:04 EliasElias 3,36855 gold badges3535 silver badges3939 bronze badges 1

• 2 @Paniyar If he doesn't understand it, he doesn't understand it. Just because it's simple for you doesn't mean it's universally simple. – mattbasta Commented Oct 13, 2010 at 13:26

Add a ment  | 

4 Answers 4

Sorted by: Reset to default 6

You can't make a function return from the function that called it in Javascript (or many other languages, afaik). You need logic in one() to do it. E.g.:

 function one() {
     return two() || "fail";
 }

function one() {
   return two();
}

You could do it, using a try-catch Block, if your function one anticipates a probable non-local-return as well as function two like this using exceptions:

function two() {
  throw {isReturn : true, returnValue : "success"}
}


function one () {
  try {
    two()
  } catch(e) {
    if(e.isReturn) return e.returnValue;
  }
  return "fail";
 }

, I believe.

function one() {    
    return two();
}