Why does my Go pointer receiver not cause an update? - go

Question

I'd like some help please on how Go pointer receivers work.
I have a contained example below of a binary search tree which hopefully helps me explain.
package main
import "fmt"
type Node struct {
key int
left, right *Node
}
func NewNode(key int) *Node {
return &Node{key, nil, nil}
}
type BST struct {
root *Node
}
func NewBinarySearchTree() *BST {
return &BST{nil}
}
func (t *BST) Insert(key int) {
if t.root == nil {
t.root = NewNode(key)
return
}
var node = t.root
for {
if key < node.key {
if node.left == nil {
node.left = NewNode(key)
return
} else {
node = node.left
}
} else {
if node.right == nil {
node.right = NewNode(key)
return
} else {
node = node.right
}
}
}
}
func inorder(node *Node) {
if node == nil {
return
}
inorder(node.left)
fmt.Print(node.key, " ")
inorder(node.right)
}
func main() {
tree := NewBinarySearchTree()
tree.Insert(3)
tree.Insert(1)
tree.Insert(2)
tree.Insert(4)
inorder(tree.root) // 1 2 3 4
}
After I wrote this, however, I thought I could simplify my insert function as follows:
func (t *BST) Insert2(key int) {
var node *Node
node = t.root
for node != nil {
if key < node.key {
node = node.left
} else {
node = node.right
}
}
node = NewNode(key)
}
However, doing it this way the tree is never updated.
My thinking was...
on the first insert the root node will be nil.
so the local variable node which references t.root will also be nil
the for loop will therefore be skipped.
node = NewNode(key) will have the same effect as t.root =
NewNode(key)
Where does my Insert2 method go wrong? Is there a way it can be tweaked?

Answer 1

You seem to be confusing the usage of pointers.
When you do node = t.root, you merely makes node point to whatever t.root points to.
Later on, when you do node = NewNode(key), you make node points to a newly created item, which is not what you wanted; you want to make t.root point to that new item instead.
Since you intend to modify variables which are of type *Node (root, left and right), we need a pointer to them, so a variable of type **Node, with one more level of indirection.
You can start by making node point to the address of t.root, node := &t.root, then you proceed to your loop.
You can try something like the following:
func (t *BST) Insert3(key int) {
node := &t.root
for *node != nil {
if key < (*node).key {
node = &(*node).left
} else {
node = &(*node).right
}
}
*node = NewNode(key)
}
Pay attention that we use the indirection operator * to access the referenced data, when checking the address on the loop, and also the key.
In the end of the function, *node = NewNode(key) does what you intended to do originally; you are assigning the newly created item to the root, left or right pointers.

Answer 2

node = NewNode(key)
That line doesn't change the tree. That line changes the local variable node; after this line, node points to a different Node, but the object it used to point to is unaffected. To insert into the tree, you have to assign to t.root, node.left, or node.right.