Trying to pass parameter by ref using list doesn't work?

Question

The function update in the following code tries to update a tree in place. Since Python doesn't support pass by ref so the function update has a parameter of list (https://realpython.com/python-pass-by-reference/).

However, n was not updated after called update([n])?

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def update(ns):
    ns[0] = ns[0].left

n = TreeNode(1, TreeNode(2), TreeNode(3))
update([n])
# n is not updated, n should be replaced by the left child of the n (2).

"Since Python doesn't support pass by ref" - any reference? Have you tried passing `n` directly? — Jan Stránský, Sep 01 '20 at 22:53
Python always passes by object reference, why do you think it doesn't? You don't need to use a list at all. — Random Davis, Sep 01 '20 at 22:54
@JanStránský "officially", Python passes "references by value" (or "by assignment"), but only for mutable objects. This has been heavily discussed, for example https://stackoverflow.com/questions/986006/how-do-i-pass-a-variable-by-reference — DeepSpace, Sep 01 '20 at 22:55
@DeepSpace I was more interested in the OP's source. Passing `n` directly to `update` and modifying it inside the function would change it in-place, as OP wants. Because it is passed "by ref" — Jan Stránský, Sep 01 '20 at 23:00
@JanStránský OP's situation is tricky because they want to change the object itself, not an attribute on it. Doing `def update(local_n): local_n = local_n.left` and calling `update(n)` will not modify the object that was passed to `update`, it will just create a new local variable `local_n` and dispose it, leaving the original unchanged — DeepSpace, Sep 01 '20 at 23:02
working with trees is tricky by default :-) If the requirement is to modify in place, then OP should modify the tree in place. @ca9163d9, could you please provide expected output after `update`? — Jan Stránský, Sep 01 '20 at 23:06
@JanStránský a TreeNode with val of 2 and no child. I've updated the question — ca9163d9, Sep 01 '20 at 23:15

score 2 · Answer 1 · answered Sep 01 '20 at 22:54

2

n = TreeNode(1, TreeNode(2), TreeNode(3))
update([n])
# n is not updated

No, n is not updated.

What's updated is [n], but you don't see it because you didn't keep a reference to it before passing it to update().

If you use

ns = [n]
update(ns)

you should see that ns has been changed.

answered Sep 01 '20 at 22:54

mkrieger1

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Right. I'd emphasis that after the update, `ns` must be inspected, not `n`. Doing `ns = [n] ; print(n) ; update(ns) ; print(n)` will not show an update, but `ns = [n] ; print(ns[0]) ; update(ns) ; print(ns[0])` will. – DeepSpace Sep 01 '20 at 22:58
No, I tried `ns = [n]` and then `update(ns)`, `n` was still not updated. – ca9163d9 Sep 01 '20 at 23:03
@ca9163d9 See my comment above. I explained what you should do – DeepSpace Sep 01 '20 at 23:04
@DeepSpace, yes, `ns[0]` will got the updated value. – ca9163d9 Sep 01 '20 at 23:12

score 2 · Accepted Answer · answered Sep 01 '20 at 23:20

ns[0] and ns[0] = xxx are synthatic sugar in python:

ns[0] means item 0 of s and internaly calls ns.__getitem__(0)

ns[0] = xxx means item 0 of ns is replaced by xxx and internaly calls ns.__setitem__(0, xxx).

It is different than writing myvar = xxx.

so in your update function, you are only mutating the list ns which is only visible inside the function.

score 1 · Answer 3 · answered Sep 01 '20 at 23:27

To update the tree in place, one option is to overwrite the attributes

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def update(n): # passed by reference!
    m = n.left # m ... node to be copied
    n.val,n.left,n.right = m.val,m.left,m.right # overwriting n attributes by m attributes

n = TreeNode(1, TreeNode(2), TreeNode(3))

print(n)
print(n.val,n.left,n.right) # original node
update(n)
print(n)
print(n.val,n.left,n.right) # changed node (but still the same instance)

score 1 · Answer 4 · answered Sep 01 '20 at 23:35

In Python, variables are labels that map to a memory reference. a=b just remaps the reference for a to the reference for b. So what you're trying to do with lists shenanigans is literally what Python does automatically with the assignment operator:

def update(ns):
    return ns.left

n = TreeNode(1, TreeNode(2), TreeNode(3))
n = update(n)

Trying to pass parameter by ref using list doesn't work?

4 Answers4