Removing an item in a Linked List with Go - go

Question

I want to remove a Node from a linked list in Go, and I have this struct and these methods:
type Node struct {
Next *Node
Val int
}
func (n *Node) Append(val int) {
end := &Node{Val: val}
here := n
for here.Next != nil {
here = here.Next
}
here.Next = end
}
func Remove(n *Node, val int) *Node {
head := n
for head.Next != nil {
if head.Next.Val == val {
head.Next = head.Next.Next
return head
}
head = head.Next
}
return head
}
func NewNode(val int) *Node {
return &Node{Val: val}
}
I want to remove an item like this:
n := NewNode(1)
n.Append(2)
n.Append(3)
n.Append(4)
n.Append(5)
m := Remove(n, 3)
for m != nil {
fmt.Println(n.Val)
m = m.Next
}
The items that get printed out are 3 and 5, not 1,2,4and5`. I re-implemented this code in Python and got the expected answer. What is going on in Go? I have a feeling it has to do something with pointers.

Answer 1

You lose the head from returning a node you use to traverse. Also you are printing out the wrong object
type Node struct {
Next *Node
Val int
}
func (n *Node) Append(val int) {
end := &Node{Val: val}
here := n
for here.Next != nil {
here = here.Next
}
here.Next = end
}
func Remove(n *Node, val int) *Node {
traverser := n
for traverser.Next != nil {
if traverser.Next.Val == val {
traverser.Next = traverser.Next.Next
return n
}
traverser = traverser.Next
}
return n
}
func NewNode(val int) *Node {
return &Node{Val: val}
}
func main() {
n := NewNode(1)
n.Append(2)
n.Append(3)
n.Append(4)
n.Append(5)
m := Remove(n, 3)
for m != nil {
fmt.Println(m.Val)
m = m.Next
}
}