Why do variable in an inner function return nan when there is the same variable name at the inner function...
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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
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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 12 mins ago
Nick Parsons
10.3k2926
asked 21 mins ago
ClaudeClaude
406
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 12 mins ago
Nick Parsons
10.3k2926
asked 21 mins ago
ClaudeClaude
406
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 12 mins ago
Nick Parsons
10.3k2926
asked 21 mins ago
ClaudeClaude
406
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 12 mins ago
Nick Parsons
10.3k2926
asked 21 mins ago
ClaudeClaude
406
edited 12 mins ago
Nick Parsons
10.3k2926
asked 21 mins ago
ClaudeClaude
406
edited 12 mins ago
Nick Parsons
10.3k2926
edited 12 mins ago
Nick Parsons
10.3k2926
edited 12 mins ago
Nick Parsons
10.3k2926
10.3k2926
asked 21 mins ago
ClaudeClaude
406
asked 21 mins ago
ClaudeClaude
406
asked 21 mins ago
ClaudeClaude
406
406
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 2 mins ago
Shidersz
9,3112933
answered 15 mins ago
jrojro
528111
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 13 mins ago
Jack BashfordJack Bashford
13.7k31848
add a comment |
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
answered 12 mins ago
Darshit ShahDarshit Shah
73
2
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
5 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
3 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 2 mins ago
Shidersz
9,3112933
answered 15 mins ago
jrojro
528111
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 2 mins ago
Shidersz
9,3112933
answered 15 mins ago
jrojro
528111
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 2 mins ago
Shidersz
9,3112933
answered 15 mins ago
jrojro
528111
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 2 mins ago
Shidersz
9,3112933
answered 15 mins ago
jrojro
528111
edited 2 mins ago
Shidersz
9,3112933
edited 2 mins ago
Shidersz
9,3112933
edited 2 mins ago
Shidersz
9,3112933
9,3112933
answered 15 mins ago
jrojro
528111
answered 15 mins ago
jrojro
528111
answered 15 mins ago
jrojro
528111
528111
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 13 mins ago
Jack BashfordJack Bashford
13.7k31848
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 13 mins ago
Jack BashfordJack Bashford
13.7k31848
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 13 mins ago
Jack BashfordJack Bashford
13.7k31848
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 13 mins ago
Jack BashfordJack Bashford
13.7k31848
answered 13 mins ago
Jack BashfordJack Bashford
13.7k31848
answered 13 mins ago
Jack BashfordJack Bashford
13.7k31848
answered 13 mins ago
Jack BashfordJack Bashford
13.7k31848
13.7k31848
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 12 mins ago
Darshit ShahDarshit Shah
73
2
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
5 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
3 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 12 mins ago
Darshit ShahDarshit Shah
73
2
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
5 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
3 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 12 mins ago
Darshit ShahDarshit Shah
73
var a=0;
function outer(){
a=2;
function inner(){
a=a+1;
console.log(a)
a = 8
}
inner()
}
outer()
answered 12 mins ago
Darshit ShahDarshit Shah
73
answered 12 mins ago
Darshit ShahDarshit Shah
73
answered 12 mins ago
Darshit ShahDarshit Shah
73
answered 12 mins ago
Darshit ShahDarshit Shah
73
73
2
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
5 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
3 mins ago
add a comment |
2
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
5 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
3 mins ago
2
2
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
5 mins ago
How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?
– Shidersz
5 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
3 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
3 mins ago
add a comment |
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