Convert singly linked list into an array using Go (Golang)

Overview

Convert singly linked list into an array using Golang

Input singly linked list: 

 "A" -> "B" -> "C" -> "D"

Output array: 

["A", "B", "C", "D"]

Program

package main
import "fmt"
type node struct {
data string
next *node
}
type singlyLinkedList struct {
len int
head *node
}
func initList() *singlyLinkedList {
return &singlyLinkedList{}
}
func (s *singlyLinkedList) AddFront(data string) {
node := &node{
data: data,
}
if s.head == nil {
s.head = node
} else {
node.next = s.head
s.head = node
}
s.len++
return
}
func (s *singlyLinkedList) Size() int {
return s.len
}
func (s *singlyLinkedList) Traverse() error {
if s.head == nil {
return fmt.Errorf("TraverseError: List is empty")
}
current := s.head
for current != nil {
fmt.Println(current.data)
current = current.next
}
return nil
}
//Function to convert singly linked list to an array
func (s *singlyLinkedList) ToArray() []string {
var myarr []string
current := s.head
for current.next != nil {
fmt.Printf("\nAdding Element to array: %s", current.data)
myarr = append(myarr, current.data)
current = current.next
}
fmt.Printf("\nAdding Element to array: %s", current.data)
myarr = append(myarr, current.data)
return myarr
}
func main() {
singleList := initList()
fmt.Printf("AddFront: D\n")
singleList.AddFront("D")
fmt.Printf("AddFront: C\n")
singleList.AddFront("C")
fmt.Printf("AddFront: B\n")
singleList.AddFront("B")
fmt.Printf("AddFront: A\n")
singleList.AddFront("A")
fmt.Printf("Size: %d\n", singleList.Size())
err := singleList.Traverse()
if err != nil {
fmt.Println(err.Error())
}
myarr := singleList.ToArray()
fmt.Println()
fmt.Println(myarr)
}

Output:

AddFront: D
AddFront: C
AddFront: B
AddFront: A
Size: 4
A
B
C
D
Adding Element to array: A
Adding Element to array: B
Adding Element to array: C
Adding Element to array: D
[A B C D]