Closure not working outside of the function scope

Question

Just started to learn closures and wrote this code in FireBug:

var later;

function outerFunc() {
  var innerVar = "Inside Outer";
  function innerFunc() {
     console.log(innerVar);
  }

  later = innerFunc;
};

outerFunc();
later();

But on calling later(); it returns an undefined value. Shouldn't it print out "Inside Outer" ?

It does for me: http://jsfiddle.net/sMuNV/ . `console.log` doesn't `return` anything, and your function doesn't `return` anything. Did you really mean to say "return"? — Ian, Apr 08 '13 at 18:26
When I run it (in Chrome's web inspector) I do get "Inside Outer" (which, to answer your question, is the correct behavior). The result of that entire expression is `undefined` because there's no return values and the last expression doesn't evaluate to anything. Are you sure "Inner Outer" isn't being printed as well? — jmar777, Apr 08 '13 at 18:26
I checked code and it works... http://jsfiddle.net/GKDev/Y3fdV/ — Givi, Apr 08 '13 at 18:28
[This](http://stackoverflow.com/questions/111102/how-do-javascript-closures-work) SO question has some great answers that may help you in the process of learning how closures work. — excentris, Apr 08 '13 at 18:34
well I mean shouldn't it print it TWO times? one time when I am calling outerFunc... one time when I am calling later() ? — , Apr 08 '13 at 18:47

score 1 · Accepted Answer · answered Apr 08 '13 at 18:30

1

var later;

function outerFunc() {
  var innerVar = "Inside Outer";
  function innerFunc() {
     console.log(innerVar);
  }

  later = innerFunc;
};

outerFunc(); //Don't run innerFunc, just set later = innerFunc
later(); // Execute innerFunc

So it logs it just one time.. It seems nice..

answered Apr 08 '13 at 18:30

fernandosavio

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well I mean shouldn't it print it TWO times? one time when I am calling outerFunc... one time when I am calling later() ? – Apr 08 '13 at 18:48
2

@user1899082 no should not, because when you call **outerFunc();** it's just assign name of **function innerFunc** to **later** and that is all. If you want to call it twice you should, instead of ***later = innerFunc;*** do so ***later = innerFunc();*** – Givi Apr 08 '13 at 18:56

Closure not working outside of the function scope

1 Answers1