const arr=["hello","world"]; for(let i=0;i<arr.length;i++) { const elem=arr[i]; console .log(elem ); } //hello //world?

Question

As we know const values cannot be changed so in this example how the const elem store "hello" and then store "world"? Any explanation please?

Answer 1

The value of a constant cannot change through reassignment, and it can't be redeclared. But it can be assigned once only during declaration.

Answer 2

After thinking and searching i got the answer, but if one feeling confused like I was
In JavaScript, const only means that the binding (the association between variable name and variable
value) is immutable. The value itself may be mutable.
In this example const elem=arr[i]; cannot equal to any other value but the value it self arr[i]; may be changed to different values.

Answer 3

Gordon I think constant equals to the looping of i so when i equals 1 it stores "one" and when i is not equal to 1 it stores "two"

Answer 4

Actually it's not reassignment

const arr is declared once only.

For const elem
It is declared in the block scope of the for loop

And garbage collector has collected the const elem for i ==0

And at i==1,
There is no const elem yet. So we can declare const elem

Answer 5

Good that you searched. But no, you are not getting closer to the truth.

arr[0] is fixed as "hello" and arr[1] is fixed as "world"

To be more clear, I used "one" and "two" in this demo for you:

https://code.sololearn.com/WGtvott81wbZ/?ref=app

I hope the above demo wipes out your misunderstanding

Can I know which reference site you are looking at? So I can point out how wrong that site is.

A better resource is Mozilla document:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/const

Quote: "Constants are block-scoped"

When you run the for loop,
for each i,
it is a different block scope.
So it may looks like that the constant is declared for more than one time.
But actually in each block scope, it is declared once only.

Answer 6

Hafsa Hussein

1. You are absolutely correct regarding the ternary operator.

2. And from console log, you see the output is
one
two
Isn't it?

3. That's because when my for loop runs, there are two i. 1 and 2.

4. In the block scope of i ==1,
constant is declared and assigned for the first time as "one"

5. In the block scope of i==2,
constant is declared and assigned, again the -->FIRST <-- time, as "two"

Answer 7

Why do you delete the comment on my demo code. I typed a long reply and it is gone because you deleted your comment 😅