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

8

1

This question already has an answer here:

• 
  Why does shadowed variable evaluate to undefined when defined in outside scope?
  
  6 answers
  
  


• 
  'Hoisted' JavaScript Variables
  
  3 answers
  
  

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 yesterday

Nick Parsons

10.3k2926

asked yesterday

ClaudeClaude

556

marked as duplicate by Matsemann, Bergi javascript
Users with the  javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

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StackExchange.helpers.removeMessages();

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yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

8

1

This question already has an answer here:

• 
  Why does shadowed variable evaluate to undefined when defined in outside scope?
  
  6 answers
  
  


• 
  'Hoisted' JavaScript Variables
  
  3 answers
  
  

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 yesterday

Nick Parsons

10.3k2926

asked yesterday

ClaudeClaude

556

marked as duplicate by Matsemann, Bergi javascript
Users with the  javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

This question already has an answer here:

• 
  Why does shadowed variable evaluate to undefined when defined in outside scope?
  
  6 answers
  
  


• 
  'Hoisted' JavaScript Variables
  
  3 answers
  
  

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 yesterday

Nick Parsons

10.3k2926

asked yesterday

ClaudeClaude

556

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 yesterday

Nick Parsons

10.3k2926

asked yesterday

ClaudeClaude

556

share|improve this question

edited yesterday

Nick Parsons

10.3k2926

asked yesterday

ClaudeClaude

556

edited yesterday

Nick Parsons

10.3k2926

edited yesterday

Nick Parsons

10.3k2926

asked yesterday

ClaudeClaude

556

asked yesterday

ClaudeClaude

556

4 Answers
4

active

oldest

votes

13

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 yesterday

Shidersz

9,3762933

answered yesterday

jrojro

604215

add a comment | 

2

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 yesterday

Jack BashfordJack Bashford

14.1k31848

add a comment | 

0

Javascript has the mechanism called hoisting .With the help of this mechanism javascript will automatically take all the variable definitions and kept it in the top of the function (whether it may be outer function or inner function). So in your case it will take only the variable definition and hoist it in the top . After hoisting you code will be like this

function outer() 
 var a = 2;

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

outer()

you are trying to increment a undefined variable . This is the reason you are getting NaN error.
So if you initialize the variable after declaration(hoisting) then you will get the result.

share|improve this answer

answered yesterday

SuRaSuRa

12114

add a comment | 

-2

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

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

inner()

outer()

share|improve this answer

edited yesterday

answered yesterday

Darshit ShahDarshit Shah

33

• 
  
  
  

  
  3
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanations and give an indication of what limitations and assumptions apply. From Review
  
  – double-beep
  yesterday
  

  
  
  
  

add a comment | 

13

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 yesterday

Shidersz

9,3762933

answered yesterday

jrojro

604215

add a comment | 

13

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 yesterday

Shidersz

9,3762933

answered yesterday

jrojro

604215

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 yesterday

Shidersz

9,3762933

answered yesterday

jrojro

604215

function outer() 
 var a=2;

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

 inner();


outer();

share|improve this answer

edited yesterday

Shidersz

9,3762933

answered yesterday

jrojro

604215

edited yesterday

Shidersz

9,3762933

edited yesterday

Shidersz

9,3762933

answered yesterday

jrojro

604215

answered yesterday

jrojro

604215

2

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 yesterday

Jack BashfordJack Bashford

14.1k31848

add a comment | 

2

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 yesterday

Jack BashfordJack Bashford

14.1k31848

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 yesterday

Jack BashfordJack Bashford

14.1k31848

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 yesterday

Jack BashfordJack Bashford

14.1k31848

answered yesterday

Jack BashfordJack Bashford

14.1k31848

answered yesterday

Jack BashfordJack Bashford

14.1k31848

0

Javascript has the mechanism called hoisting .With the help of this mechanism javascript will automatically take all the variable definitions and kept it in the top of the function (whether it may be outer function or inner function). So in your case it will take only the variable definition and hoist it in the top . After hoisting you code will be like this

function outer() 
 var a = 2;

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

outer()

you are trying to increment a undefined variable . This is the reason you are getting NaN error.
So if you initialize the variable after declaration(hoisting) then you will get the result.

share|improve this answer

answered yesterday

SuRaSuRa

12114

add a comment | 

0

Javascript has the mechanism called hoisting .With the help of this mechanism javascript will automatically take all the variable definitions and kept it in the top of the function (whether it may be outer function or inner function). So in your case it will take only the variable definition and hoist it in the top . After hoisting you code will be like this

function outer() 
 var a = 2;

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

outer()

you are trying to increment a undefined variable . This is the reason you are getting NaN error.
So if you initialize the variable after declaration(hoisting) then you will get the result.

share|improve this answer

answered yesterday

SuRaSuRa

12114

add a comment | 

Javascript has the mechanism called hoisting .With the help of this mechanism javascript will automatically take all the variable definitions and kept it in the top of the function (whether it may be outer function or inner function). So in your case it will take only the variable definition and hoist it in the top . After hoisting you code will be like this

function outer() 
 var a = 2;

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

outer()

you are trying to increment a undefined variable . This is the reason you are getting NaN error.
So if you initialize the variable after declaration(hoisting) then you will get the result.

share|improve this answer

answered yesterday

SuRaSuRa

12114

function outer() 
 var a = 2;

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

outer()

function outer() 
 var a = 2;

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

outer()

function outer() 
 var a = 2;

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

outer()

share|improve this answer

answered yesterday

SuRaSuRa

12114

answered yesterday

SuRaSuRa

12114

answered yesterday

SuRaSuRa

12114

-2

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

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

inner()

outer()

share|improve this answer

edited yesterday

answered yesterday

Darshit ShahDarshit Shah

33

• 
  
  
  

  
  3
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanations and give an indication of what limitations and assumptions apply. From Review
  
  – double-beep
  yesterday
  

  
  
  
  

add a comment | 

-2

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

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

inner()

outer()

share|improve this answer

edited yesterday

answered yesterday

Darshit ShahDarshit Shah

33

• 
  
  
  

  
  3
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanations and give an indication of what limitations and assumptions apply. From Review
  
  – double-beep
  yesterday
  

  
  
  
  

add a comment | 

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

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

inner()

outer()

share|improve this answer

edited yesterday

answered yesterday

Darshit ShahDarshit Shah

33

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

inner()

outer()

share|improve this answer

edited yesterday

answered yesterday

Darshit ShahDarshit Shah

33

answered yesterday

Darshit ShahDarshit Shah

33

answered yesterday

Darshit ShahDarshit Shah

33