在嵌入式功能中访问功能

Question

First, I'm quite new to Go, so please forgive any lapses or errors in terminology. I suspect my lack of command with the terminology is also partially to blame for not finding an answer to the following question after many hours of looking.

In short, I would like the output of the following code to be

I am the Adult
I am the Child

Where instead the output is

I am the Adult
I am the Adult

Code:

package main

import "fmt"

type Human struct {
    age  uint
    name string
}

func (h Human) sayName() error {
    fmt.Println("I am the Adult")
    return nil
}

func (h Human) Introduce() error {
    h.sayName()
    return nil
}

type Child struct {
    Human
}

func (c Child) sayName() error {
    fmt.Println("I am the Child")
    return nil
}

func main() {
    h := Human{}
    h.Introduce()

    c := Child{Human{}}
    c.Introduce()
}

So in essence, though Introduce() is only implemented in the embedded type, Human, it calls sayName(), which is implemented in both the embedded type and the embedding type.

I understand that the current output is the way as it is because the embedded Human struct doesn't "know" it's embedded and thus could never call Child.sayName and would only call its own sayName() function.

Is there a way to instantiate a Human struct (or a struct embedding one) in which you can "replace" Human.sayName() with an alternate sayName() function?

Answer 1

The way to get this kind of late-bound behavior is to use interfaces. If Introduce was a method on an interface that required a sayName method, then Human and Child would both satisfy that interface, and could Introduce themselves, and the proper sayName method would be called in either case, because it would be dispatched via the interface type, instead of via Human.