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
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
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
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 24 mins ago
Nick Parsons
10.3k2926
10.3k2926
asked 32 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.
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();
add a comment |
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
3
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
17 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
15 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.
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.
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.
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 14 mins ago
Shidersz
9,3112933
9,3112933
answered 26 mins ago
jrojro
562112
562112
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();
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();
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();
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 24 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()
3
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
17 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
15 mins ago
add a comment |
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
3
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
17 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
15 mins ago
add a comment |
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
answered 24 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
17 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
15 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
17 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
15 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
17 mins ago
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
17 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
15 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
15 mins ago
add a comment |
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