Vwwxrf

Is it professional to write unrelated content in an almost-empty email?

How do we know the LHC results are robust?

Anatomically Correct Strange Women In Ponds Distributing Swords

What happened in Rome, when the western empire "fell"?

How does the mv command work with external drives?

How to invert MapIndexed on a ragged structure? How to construct a tree from rules?

What can we do to stop prior company from asking us questions?

How does the Z80 determine which peripheral sent an interrupt?

Is it ever safe to open a suspicious html file (e.g. email attachment)?

Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?

Sending manuscript to multiple publishers

Unreliable Magic - Is it worth it?

Is micro rebar a better way to reinforce concrete than rebar?

To not tell, not take, and not want

Why am I allowed to create multiple unique pointers from a single object?

How did people program for Consoles with multiple CPUs?

Why does standard notation not preserve intervals (visually)

In excess I'm lethal

If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?

Complex fractions

What is the result of assigning to std::vector<T>::begin()?

Return the Closest Prime Number

Are there any unintended negative consequences to allowing PCs to gain multiple levels at once in a short milestone-XP game?

How to avoid supervisors with prejudiced views?

Why do variable in an inner function return nan when there is the same variable name at the inner function declared after log

6

What's happening here? I get a different result if I declare a variable after console.log in the inner function

I understand that var has a functional scope and inner function can access the variable from their parent

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log NaN
 var a = 8
 
 inner()

outer()

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log 3
 var b = 8
 
 inner()

outer()

The log returns NaN in the first example and log 3 in the second example

javascript function

share|improve this question

edited 17 mins ago

Nick Parsons

10.3k2926

asked 26 mins ago

ClaudeClaude

426

add a comment | 

6

What's happening here? I get a different result if I declare a variable after console.log in the inner function

I understand that var has a functional scope and inner function can access the variable from their parent

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log NaN
 var a = 8
 
 inner()

outer()

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log 3
 var b = 8
 
 inner()

outer()

The log returns NaN in the first example and log 3 in the second example

javascript function

share|improve this question

edited 17 mins ago

Nick Parsons

10.3k2926

asked 26 mins ago

ClaudeClaude

426

add a comment | 

What's happening here? I get a different result if I declare a variable after console.log in the inner function

I understand that var has a functional scope and inner function can access the variable from their parent

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log NaN
 var a = 8
 
 inner()

outer()

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log 3
 var b = 8
 
 inner()

outer()

The log returns NaN in the first example and log 3 in the second example

javascript function

share|improve this question

edited 17 mins ago

Nick Parsons

10.3k2926

asked 26 mins ago

ClaudeClaude

426

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log NaN
 var a = 8
 
 inner()

outer()

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log 3
 var b = 8
 
 inner()

outer()

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log NaN
 var a = 8
 
 inner()

outer()

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log NaN
 var a = 8
 
 inner()

outer()

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log 3
 var b = 8
 
 inner()

outer()

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a) //log 3
 var b = 8
 
 inner()

outer()

share|improve this question

edited 17 mins ago

Nick Parsons

10.3k2926

asked 26 mins ago

ClaudeClaude

426

share|improve this question

edited 17 mins ago

Nick Parsons

10.3k2926

asked 26 mins ago

ClaudeClaude

426

edited 17 mins ago

Nick Parsons

10.3k2926

edited 17 mins ago

Nick Parsons

10.3k2926

asked 26 mins ago

ClaudeClaude

426

asked 26 mins ago

ClaudeClaude

426

3 Answers
3

active

oldest

votes

8

This is due to hoisting

The declaration of a in the inner function is hoisted to the top of the function, overriding the outer function's a, so a is undefined

undefined++ returns NaN, hence your result.

Your code is equivalent to:

function outer() 
 var a=2;

 function inner() 
 var a;
 a++;
 console.log(a); //log NaN
 a = 8;
 

 inner();


outer();

Rewriting your code in this way makes it easy to see what's going on.

share|improve this answer

edited 7 mins ago

Shidersz

9,3112933

answered 20 mins ago

jrojro

552111

add a comment | 

1

Because var is hoisted through the function, you're essentially running undefined++ which is NaN. If you remove var a = 8 in inner, the code works as expected:

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a);
 
 inner();

outer();

share|improve this answer

answered 17 mins ago

Jack BashfordJack Bashford

13.8k31848

add a comment | 

-1

var a=0;
function outer()
a=2;
function inner()
a=a+1;
console.log(a)
 a = 8

inner()

outer()

share|improve this answer

answered 17 mins ago

Darshit ShahDarshit Shah

53

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
  
  – Shidersz
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code
  
  – Darshit Shah
  8 mins ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55428371%2fwhy-do-variable-in-an-inner-function-return-nan-when-there-is-the-same-variable%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

8

This is due to hoisting

The declaration of a in the inner function is hoisted to the top of the function, overriding the outer function's a, so a is undefined

undefined++ returns NaN, hence your result.

Your code is equivalent to:

function outer() 
 var a=2;

 function inner() 
 var a;
 a++;
 console.log(a); //log NaN
 a = 8;
 

 inner();


outer();

Rewriting your code in this way makes it easy to see what's going on.

share|improve this answer

edited 7 mins ago

Shidersz

9,3112933

answered 20 mins ago

jrojro

552111

add a comment | 

8

This is due to hoisting

The declaration of a in the inner function is hoisted to the top of the function, overriding the outer function's a, so a is undefined

undefined++ returns NaN, hence your result.

Your code is equivalent to:

function outer() 
 var a=2;

 function inner() 
 var a;
 a++;
 console.log(a); //log NaN
 a = 8;
 

 inner();


outer();

Rewriting your code in this way makes it easy to see what's going on.

share|improve this answer

edited 7 mins ago

Shidersz

9,3112933

answered 20 mins ago

jrojro

552111

add a comment | 

This is due to hoisting

The declaration of a in the inner function is hoisted to the top of the function, overriding the outer function's a, so a is undefined

undefined++ returns NaN, hence your result.

Your code is equivalent to:

function outer() 
 var a=2;

 function inner() 
 var a;
 a++;
 console.log(a); //log NaN
 a = 8;
 

 inner();


outer();

Rewriting your code in this way makes it easy to see what's going on.

share|improve this answer

edited 7 mins ago

Shidersz

9,3112933

answered 20 mins ago

jrojro

552111

function outer() 
 var a=2;

 function inner() 
 var a;
 a++;
 console.log(a); //log NaN
 a = 8;
 

 inner();


outer();

share|improve this answer

edited 7 mins ago

Shidersz

9,3112933

answered 20 mins ago

jrojro

552111

edited 7 mins ago

Shidersz

9,3112933

edited 7 mins ago

Shidersz

9,3112933

answered 20 mins ago

jrojro

552111

answered 20 mins ago

jrojro

552111

1

Because var is hoisted through the function, you're essentially running undefined++ which is NaN. If you remove var a = 8 in inner, the code works as expected:

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a);
 
 inner();

outer();

share|improve this answer

answered 17 mins ago

Jack BashfordJack Bashford

13.8k31848

add a comment | 

1

Because var is hoisted through the function, you're essentially running undefined++ which is NaN. If you remove var a = 8 in inner, the code works as expected:

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a);
 
 inner();

outer();

share|improve this answer

answered 17 mins ago

Jack BashfordJack Bashford

13.8k31848

add a comment | 

Because var is hoisted through the function, you're essentially running undefined++ which is NaN. If you remove var a = 8 in inner, the code works as expected:

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a);
 
 inner();

outer();

share|improve this answer

answered 17 mins ago

Jack BashfordJack Bashford

13.8k31848

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a);
 
 inner();

outer();

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a);
 
 inner();

outer();

function outer() 
 var a = 2;

 function inner() 
 a++;
 console.log(a);
 
 inner();

outer();

share|improve this answer

answered 17 mins ago

Jack BashfordJack Bashford

13.8k31848

answered 17 mins ago

Jack BashfordJack Bashford

13.8k31848

answered 17 mins ago

Jack BashfordJack Bashford

13.8k31848

-1

var a=0;
function outer()
a=2;
function inner()
a=a+1;
console.log(a)
 a = 8

inner()

outer()

share|improve this answer

answered 17 mins ago

Darshit ShahDarshit Shah

53

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
  
  – Shidersz
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code
  
  – Darshit Shah
  8 mins ago
  

  
  
  
  

add a comment | 

-1

var a=0;
function outer()
a=2;
function inner()
a=a+1;
console.log(a)
 a = 8

inner()

outer()

share|improve this answer

answered 17 mins ago

Darshit ShahDarshit Shah

53

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
  
  – Shidersz
  10 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code
  
  – Darshit Shah
  8 mins ago
  

  
  
  
  

add a comment | 

var a=0;
function outer()
a=2;
function inner()
a=a+1;
console.log(a)
 a = 8

inner()

outer()

share|improve this answer

answered 17 mins ago

Darshit ShahDarshit Shah

53

var a=0;
function outer()
a=2;
function inner()
a=a+1;
console.log(a)
 a = 8

inner()

outer()

share|improve this answer

answered 17 mins ago

Darshit ShahDarshit Shah

53

answered 17 mins ago

Darshit ShahDarshit Shah

53

answered 17 mins ago

Darshit ShahDarshit Shah

53