/*
* methods in Go are dispatched statically.
* NOT inherited.
* If you define a struct as anonymous include of
* another struct, methods are inherited.
* Non-anonymous == methods are not inherited
* I never remember all of this!
*
* @author gtowell
* created: Aug 6, 2021
 */
package main

import "fmt"

// simple struct
type SNum struct {
	num int
}
// simple method on struct
func (s SNum) printer() {
	fmt.Printf("SNum num=%d\n", s.num)
}
// a new type that is just a renaming of the original struct
type SSNum SNum
// new method on the renaming
func (s SSNum) printer() {
	fmt.Printf("SSNum num=%d\n", s.num)
}

// struct with anonymous include
type SNumAndStr struct {
	SNum
	str string
}

// similar struct but with named include.
type S2 struct {
	sn SNum
	str string
}

func main() {
	sn := SNum{num:5}
	sn.printer()
	// cast SNum to SSNum
	ssn:= SSNum(sn)
	// the printer method used is from SSNUM
	// In Java it would be from SNum
	ssn.printer()

	// with anonymous include the methods on included struct are available
	nands := SNumAndStr{SNum{num:7}, "test"}
	fmt.Println(nands)
	nands.printer()

	// with named include, methods are NOT available.
	s2 := S2{SNum{9}, "hello"}
	fmt.Println(s2)
	//s2.printer()
}