Function arguments in execution context and scope in JavaScript

Question

In chapter 4 of Professional JavaScript for Web Developers by Nicholas C. Zakas, author says that function arguments follow the same access rules as any other variable in the execution context. For that, I tested the follows code:

function n1(num1, num2) {
   function n2() {
      var num3 = (num1 + num2);
      console.log(num3);
   }
}

I called n1() function with: n1(1, 2). I thought that the result would be 3, but I get undefined.

Why this behavior?

Pavlo · Accepted Answer · 2013-12-22 19:50:50Z

4

Neither n1, nor n2 doesn't return anything (that's where undefined comes from). Additionally, n2 is never called. If you want to get 3 from n1(1, 2) you have to modify it something like this:

function n1(num1, num2) {
   function n2() {
      var num3 = (num1 + num2);
      console.log(num3);
      return num3;
   }
   return n2();
}

edited Dec 22, 2013 at 19:50

answered Dec 21, 2013 at 21:02

Pavlo

45.2k14 gold badges83 silver badges114 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

leoMestizo Over a year ago

Oh! That's right! Too obvious and I couldn't see it! Thanks a lot Pavlo!

leoMestizo Over a year ago

BTW, can you modify the return statement in n2() function? It must be "num3". I tryed but the edit must be 6 characters at least.

RobG Over a year ago

@leoMestizo—edit with say 10 spurious characters, then edit again to remove them.

Collectives™ on Stack Overflow

Function arguments in execution context and scope in JavaScript

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related