Structs and interface types - go

Question

Could someone please explain (or if it's a long thing to explain at least point me to some documentation) why this code is behaving like it does. It's a bit lengthy but I couldn't figure out a good way of shortening it without losing too much info.
What confuses me is that I need to have the method "Find" on each type of struct for it to be recognized as the type it is. With Find on only the baseTypeImp is it printing that each is baseTypeImp, if I have it on baseTypeImp and advancedBaseTypeImp are those two identified correctly, but not the last type.
PlaygroundLink
My Code
package main
import (
"fmt"
"reflect"
)
type BaseType interface {
Name() string
Find(string) BaseType
Children() []BaseType
Options()
}
type baseTypeImp struct {
name string
children []BaseType
}
func (b baseTypeImp) Options() {
fmt.Println("Not implemented")
}
func (b baseTypeImp) Find(name string) BaseType {
if b.name == name {
return b
}
for _, c := range b.children {
if m := c.Find(name); m != nil {
return m
}
}
return nil
}
func (b baseTypeImp) Name() string {
return b.name
}
func (b baseTypeImp) Children() []BaseType {
return b.children
}
type AdvancedBaseType interface {
Value()
}
type advancedBaseTypeImp struct {
baseTypeImp
}
func (a advancedBaseTypeImp) Options() {
fmt.Println("Is implemented")
}
func (a advancedBaseTypeImp) Value() {
fmt.Println("Value called")
}
func (a advancedBaseTypeImp) Find(name string) BaseType {
if a.name == name {
return a
}
for _, c := range a.children {
if m := c.Find(name); m != nil {
return m
}
}
return nil
}
type reallyAdvancedBaseTypeImp advancedBaseTypeImp
func newThingy(name, variant string, children []BaseType) BaseType {
base := baseTypeImp{name: name, children: children}
switch variant {
case "advanced":
return advancedBaseTypeImp{baseTypeImp: base}
case "reallyAdvanced":
return reallyAdvancedBaseTypeImp{baseTypeImp: base}
}
return base
}
func whatType(b BaseType) {
if b == nil {
return
}
fooType := reflect.TypeOf(b)
fmt.Println(b.Name(), " is type ", fooType.Name())
b.Options()
}
func main() {
advanced := make([]BaseType, 0, 5)
for i := 0; i < 5; i++ {
advanced = append(advanced, newThingy(fmt.Sprintf("Advanced %d", i), "advanced", nil))
}
reallyAdvanced := make([]BaseType, 0, 2)
for i := 0; i < 2; i++ {
reallyAdvanced = append(reallyAdvanced, newThingy(fmt.Sprintf("ReallyAdvanced %d", i), "reallyAdvanced", advanced[i:i+3]))
}
basic := newThingy("Basic parent", "basic", reallyAdvanced)
whatType(basic.Find("Basic parent"))
for i := 0; i < 2; i++ {
whatType(basic.Find(fmt.Sprintf("Advanced %d", i)))
whatType(basic.Find(fmt.Sprintf("ReallyAdvanced %d", i)))
}
}
The output now is:
Basic parent is type baseTypeImp
Not implemented
Advanced 0 is type advancedBaseTypeImp
Is implemented
ReallyAdvanced 0 is type baseTypeImp
Not implemented
Advanced 1 is type advancedBaseTypeImp
Is implemented
ReallyAdvanced 1 is type baseTypeImp
Not implemented
What I'd like to have is:
Basic parent is type baseTypeImp
Not implemented
Advanced 0 is type advancedBaseTypeImp
Is implemented
ReallyAdvanced 0 is type reallyAdvancedBaseTypeImp
Is implemented
Advanced 1 is type advancedBaseTypeImp
Is implemented
ReallyAdvanced 1 is type reallyAdvancedBaseTypeImp
Is implemented
Is there a way to get this working without having to implement Find on each level? It kind of defeats the purpose of having the common methods in the top layer struct. And of course, if possible explain why it's behaving the way it does.

Answer 1

When you embed a struct into another one, the outer struct gets the methods of the embedded one.
type inner struct {
}
func (i inner) f() {}
type outer struct {
inner
}
...
x:=outer{}
x.f()
You can think of this as syntactic sugar for:
type outer2 struct {
i inner
}
y.i.f()
When you embed inner without a field name, you simply drop the i in the call. You can still call x.inner.f() in the first example.
When you redefine a type, it does not get the methods defined for its base type.
type newtype inner
newtype does not have f() defined. However, if inner also embeds other structs (like you do), then those are still embedded in the new type, so those function are defined:
type deepest struct {}
func (deepest) deep() {}
type inner struct{}
func (inner) in() {}
type outer inner
Above, the type outer does not have the function in, but it has deep.
So when you call reallyAdvancedBaseTypeImp.Find(), what you're really doing is reallyAdvancedBaseTypeImp.baseTypeImp.Find(), which works on that portion of the struct. That's why you're getting the types you get.
You're trying to emulate method overriding. You can't really do that. Formulate your problem in a different way.