Why do variable in an inner function return nan when there is the same variable name at the inner function...
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
edited 14 mins ago
Nick Parsons
10.3k2926
asked 23 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
edited 14 mins ago
Nick Parsons
10.3k2926
asked 23 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
edited 14 mins ago
Nick Parsons
10.3k2926
asked 23 mins ago
ClaudeClaude
426
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
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()javascript function
javascript function
edited 14 mins ago
Nick Parsons
10.3k2926
asked 23 mins ago
ClaudeClaude
426
edited 14 mins ago
Nick Parsons
10.3k2926
asked 23 mins ago
ClaudeClaude
426
edited 14 mins ago
Nick Parsons
10.3k2926
edited 14 mins ago
Nick Parsons
10.3k2926
edited 14 mins ago
Nick Parsons
10.3k2926
10.3k2926
asked 23 mins ago
ClaudeClaude
426
asked 23 mins ago
ClaudeClaude
426
asked 23 mins ago
ClaudeClaude
426
426
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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.
edited 5 mins ago
Shidersz
9,3112933
answered 17 mins ago
jrojro
552111
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();answered 15 mins ago
Jack BashfordJack Bashford
13.8k31848
add a comment |
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
answered 15 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
8 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
6 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
edited 5 mins ago
Shidersz
9,3112933
answered 17 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.
edited 5 mins ago
Shidersz
9,3112933
answered 17 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.
edited 5 mins ago
Shidersz
9,3112933
answered 17 mins ago
jrojro
552111
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.
edited 5 mins ago
Shidersz
9,3112933
answered 17 mins ago
jrojro
552111
edited 5 mins ago
Shidersz
9,3112933
edited 5 mins ago
Shidersz
9,3112933
edited 5 mins ago
Shidersz
9,3112933
9,3112933
answered 17 mins ago
jrojro
552111
answered 17 mins ago
jrojro
552111
answered 17 mins ago
jrojro
552111
552111
add a comment |
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();answered 15 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();answered 15 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();answered 15 mins ago
Jack BashfordJack Bashford
13.8k31848
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();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();answered 15 mins ago
Jack BashfordJack Bashford
13.8k31848
answered 15 mins ago
Jack BashfordJack Bashford
13.8k31848
answered 15 mins ago
Jack BashfordJack Bashford
13.8k31848
answered 15 mins ago
Jack BashfordJack Bashford
13.8k31848
13.8k31848
add a comment |
add a comment |
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
answered 15 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
8 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
6 mins ago
add a comment |
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
answered 15 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
8 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
6 mins ago
add a comment |
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
answered 15 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()
answered 15 mins ago
Darshit ShahDarshit Shah
53
answered 15 mins ago
Darshit ShahDarshit Shah
53
answered 15 mins ago
Darshit ShahDarshit Shah
53
answered 15 mins ago
Darshit ShahDarshit Shah
53
53
3
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
8 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
6 mins ago
add a comment |
3
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
8 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
6 mins ago
3
3
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
8 mins ago
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
8 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
6 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
6 mins ago
add a comment |
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