从没有转换类型的继承结构执行实际对象方法到继承类型

package main

import (
    "fmt"
)

type IA interface {
    Parse()
    Name() string
}

type A struct {
    IA
}

func (a *A) Name() string {
    return "AName"
}

func (a *A) Parse() {
    fmt.Println("A-" + a.Name())
}

type B struct {
    A
}

func (b *B) Name() string {
    return "BName"
}


func main() {
    a := &A{}
    b := &B{}

    a.Parse()
    b.Parse() // I would like to see "A-BName"
}

Playground

When I execute method from inherited struct with execute another struct method in it - is executed method from inherited struct, not actual object type.

As JimB said, there is no proper inheritance in Go as in other languages. To help visualize this, let's take your code in main():

a := &A{}
b := &B{}

a.Parse()
b.Parse()

When a.Parse() is called, the Go compiler checks if a has a method Parse(), and it does. So it calls the method a.Parse.

When b.Parse() is called, the Go compiler checks if b has a method Parse(), and it does not - but an embedded field inside the struct does! So it simply uses the method from that field, and thus b.A.Parse() is called. In fact, you could change B to this:

type B struct {
    A A
}

And the only difference would be that b.Parse() wouldn't work and that you'd have to specify .A manually (thus b.A.Parse()), but it remains functionally the same.