When using 'define var2 var1', both variables get linked

[Drogebot]

When using this code:

(define var1 '((1 #f #f) (2 #f #f) (3 #f #f)))
(define var2 var1)
(set-nth! (nth var2 2) 2 #t)

The value of the variables after this code is:
var1 = ((1 #f #f) (2 #t #f) (3 #f #f)))
var2 = ((1 #f #f) (2 #t #f) (3 #f #f))) //notice how they both have a #t
But what I would expect the two variables to be is:
var1 = ((1 #f #f) (2 #f #f) (3 #f #f)))
var2 = ((1 #f #f) (2 #t #f) (3 #f #f))) //notice how only var2 has a #t

So basically the two variables get linked, preventing me from making more complex code.
Is this intended and is there a workaround or does this need fixing?

[ToadKing]

Lists are mutable objects so if you want to modify one while leaving the other alone you should make a copy of it first.

(define var2 (copy var1))

However, it might make sense for define to do this copy implicitly. I'll discuss it with the rest of the team.

[Drogebot]

ahh, yes that works perfectly! Thanks.

[dastels]

It currently has the correct behaviour. Var2 gets bound to the value of var1. That is a list. So var1 and var2 are bith bound to the same cons cell ie the same list. So if you modify the list it's the same list for both.

But you should try to avoid mutating lists. Or anything, really.