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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

7

1

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 1 hour ago

Nick Parsons

10.3k2926

asked 1 hour ago

ClaudeClaude

476

add a comment | 

7

1

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 1 hour ago

Nick Parsons

10.3k2926

asked 1 hour ago

ClaudeClaude

476

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 1 hour ago

Nick Parsons

10.3k2926

asked 1 hour ago

ClaudeClaude

476

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 1 hour ago

Nick Parsons

10.3k2926

asked 1 hour ago

ClaudeClaude

476

share|improve this question

edited 1 hour ago

Nick Parsons

10.3k2926

asked 1 hour ago

ClaudeClaude

476

edited 1 hour ago

Nick Parsons

10.3k2926

edited 1 hour ago

Nick Parsons

10.3k2926

asked 1 hour ago

ClaudeClaude

476

asked 1 hour ago

ClaudeClaude

476

3 Answers
3

active

oldest

votes

11

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 1 hour ago

Shidersz

9,3362933

answered 1 hour ago

jrojro

584115

add a comment | 

3

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 1 hour ago

Jack BashfordJack Bashford

13.8k31848

add a comment | 

-1

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 38 mins ago

answered 1 hour 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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  41 mins ago
  

  
  
  
  

add a comment | 

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11

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 1 hour ago

Shidersz

9,3362933

answered 1 hour ago

jrojro

584115

add a comment | 

11

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 1 hour ago

Shidersz

9,3362933

answered 1 hour ago

jrojro

584115

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 1 hour ago

Shidersz

9,3362933

answered 1 hour ago

jrojro

584115

function outer() 
 var a=2;

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

 inner();


outer();

share|improve this answer

edited 1 hour ago

Shidersz

9,3362933

answered 1 hour ago

jrojro

584115

edited 1 hour ago

Shidersz

9,3362933

edited 1 hour ago

Shidersz

9,3362933

answered 1 hour ago

jrojro

584115

answered 1 hour ago

jrojro

584115

3

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 1 hour ago

Jack BashfordJack Bashford

13.8k31848

add a comment | 

3

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 1 hour 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 1 hour 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 1 hour ago

Jack BashfordJack Bashford

13.8k31848

answered 1 hour ago

Jack BashfordJack Bashford

13.8k31848

answered 1 hour ago

Jack BashfordJack Bashford

13.8k31848

-1

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 38 mins ago

answered 1 hour 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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  41 mins ago
  

  
  
  
  

add a comment | 

-1

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 38 mins ago

answered 1 hour 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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  41 mins ago
  

  
  
  
  

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 38 mins ago

answered 1 hour 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

edited 38 mins ago

answered 1 hour ago

Darshit ShahDarshit Shah

53

answered 1 hour ago

Darshit ShahDarshit Shah

53

answered 1 hour ago

Darshit ShahDarshit Shah

53