Resolved How to duplicate a list in a new one?

Hi,

When I do this:


I don't get 2 lists, but 2 variables pointing on the same list. Modifying one variable will modify the second one.

How do I get 2 different lists, so I can modify the first list without modifying the second list?

Do I have to use For Each Loop necessarily ?
If yes, then do I have to copy all the fields of each item of the list A to all the field of each item of the list B?

Thanks!

 

im sorry bud i don't understand
i kinda do but..

you just want to change one list but not the other depending on a value?
just don't reference it then? i dont understand sorry man.

clearly you're trying to sort an inventory
usually i'd use the 'for each node'

 

Let me explain it differently:

Usually, when I do:

Set var a = 5
Set var b = var a

Then I change only var a:
Set var a = 10

Result is:
var a = 10
var b = 5

But in the case of list, the result is:
var a = 10
var b = 10.

My vague understanding is that for some types, like lists in this case, the instruction Set does not put the contenant of var a in var b, but make the var b and var a point on the same data in memory. Thus, var a, and var b will always be the same. That is not what I want.

I'd like to have no link between var a and var b, So once I've copied var a in var b, modifying var a won't affect var b and vice versa.

 

How do you do that?

 

seems like an order of operations thing.

if you're doing

float a = 5;
float b = a;
// a would be 5 here

//then
b = 10;

// b is 10 and a is 5 here

 

No it's not an order of operation issue.

Both variables are the same, changing one will change the other one.

The Set instruction is not assigning values, but an object reference. So both variables reference the same object.

 

like this is the reference of the list item
if you don't directly reference a list, it can't change right

 

I have this list A:


I want to create a copy of this list in real time to create list B:

If I use Set List B = List A
And I modify list B, it modifies also list A.
However, what I want is to modify list B content without modifying list A content. how would you do that?

 

And doing this does not solve the issue:


Modifying the new list will still modify the initial list...

 

In your example, the 2 lists already exist, so they are independent.
In my case the second list does not exit. It is created in real time as a copy of the first one.

And when it is a copy, they both have the same reference object, so modifying the first will modify the second.

 

beauty

okay what you would want to do is generate a new temp list from that list like this.
you can see how it copies the data on runtime and adds to an empty list




you could then save that runtime list in another instance of the list



after it runs we can see the new data that was saved

 

The saving part is not required. I need to modify Temp list without modifying the OG list.
How would you do that?

Did you try to modify Temp list and see what happened to OG List?

 

Yes, taking the order of operations into consideration, we just hook this between the two to change the value before it's saved.

the result is an unchanged OG list
while the Templist is updated with the new string item










because we're generating an entirely new list and referencing it directly, only that list can change values while the OG is still OG.

 

I set this script. List b is empty before On Start is triggered.



On start, the copy is made:


So far so good.

Now I connect the modification part applied only to List b:


On start I get this:



Why the list a is modified?!!!


BTW, this works:



But why?!

 

because you're setting it like that once the loop is done

 

No but i am asking why modifying list b modifies list a.

I am talking about my 3rd screenshot.