CODEWITHSUNDEEP
go
W.Cameron
W.Cameron

Reputation: 124

Can go convert an interface type pointer to VALUE in passing function parameter?

When I studied Go by "Go In Action" book, there is a segment code like below.

I haven't learned Go too much in its interface concept.

The segment defines a interface called notifier used to notify people, and two types (user , admin) impl this interface.There is a middleware, says sendNotification(), that uniforms the type who impls notifier interface

What confuses me is the parameter of middleware function sendNotification(), it receives a value (not pointer!) of inferface type(notifier) .But, why uses a pointer to invoke this function in main function:

sendNodification(&lisa)

sendNodification(&bil)

?? just why?

windows 10. Go 1.12.2

package main

import "fmt"

type notifier interface {nodify()}

type user struct {name string;email string}

type admin struct{name string;email string}

func (u *user) nodify (){
    fmt.Printf("sending user email to %s<%s>\n", u.name, u.email)
}
func (u *admin) nodify(){
    fmt.Printf("sending admin email to %s<%s>\n", u.name, u.email)
}

func sendNodification(n notifier){
    n.nodify()
}

func main(){
    bil := user{"Bill", "[email protected]"}
    sendNodification(&bil)
    lisa := admin{"Lisa", "[email protected]"}
    sendNodification(&lisa)
}

Upvotes: 0

Views: 98

Answers (1)

Adrian
Adrian

Reputation: 46442

Because the methods are declared with pointer receivers (func (u *user) nodify rather than func (u user) nodify), it's the pointer to the type that implements the interface. So you can pass a *user to a function accepting a notifier because *user implements the interface.

Upvotes: 3

Related Questions