The following code fails, because after using B.Assign(A), the information on the specific type of B is lost (at least that's what I think goes wrong here):
package main
import "fmt"
type Methods interface {
Method()
Assign(Methods)
Set(Methods)
}
type Parent struct {
Methods
assigned Methods
}
type ChildA struct {
Parent
}
type ChildB struct {
Parent
msg string
}
func (this *Parent) Assign(other Methods) {
//Some other logic here
other.Set(this)
}
func (this *Parent) Set(other Methods) {
this.assigned = other
}
func (c *ChildA) Method() {
fmt.Println("I'm child A")
}
func (c *ChildB) Method() {
fmt.Println("I'm child B, and I approve this message:", c.msg)
}
func main() {
A := new(ChildA)
B := new(ChildB)
B.msg = "my message"
B.Assign(A)
A.assigned.Method()
}
Now, in order to avoid this, I would have to make another method, that has exactly the same definition as Parent.Assign(), but different receiver:
func (this *Parent) Assign(other Methods) {
//Some other logic here
other.Set(this)
}
func (this *ChildB) Assign(other Methods) {
//Same as above
other.Set(this)
}
This is rather ugly. Is there a way to preserve the information about B's type when calling the method from it's embedded struct Parent, without duplicating the code?
Is there a way to preserve the information about B's type when calling the method from it's embedded struct Parent, without duplicating the code?
No. When you call the embedded method, it is called with a pointer to the embedded struct.
Unfortunately as attractive as it first seems, embedded objects don't make Go into an object oriented language.
You might be better off just having one type with function pointers for the implementation.