golang method pointer to receiver

Question

I have the following struct and :

type Person struct {
    Name    string
}


steve := Person{Name: "Steve"}

Can you explain how the following 2 methods (one without the pointer and one with in the receiver) both are able to print the p.Name?

func (p *Person) Yell() {
    fmt.Println("Hi, my name is", p.Name)
}

func (p Person) Yell(){
    fmt.Println("YELLING MY NAME IS", p.Name)
}

steve.Yell()

Wouldn't the Name not exist when pointing straight to the Person (not the instance steve?)

OneOfOne · Accepted Answer · 2014-06-10 02:36:27Z

4

Both point to the instance, however (p Person) points to a new copy every time you call the function, where (p *Person) will always point to the same instance.

Check this example :

func (p Person) Copy() {
    p.Name = "Copy"
}

func (p *Person) Ptr() {
    p.Name = "Ptr"
}

func main() {
    p1, p2 := Person{"Steve"}, Person{"Mike"}
    p1.Copy()
    p2.Ptr()
    fmt.Println("copy", p1.Name)
    fmt.Println("ptr", p2.Name)
}

Also read Effective Go, it's a great resource to the language.

answered Jun 10, 2014 at 2:36

OneOfOne

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2 Comments

lukemh Over a year ago

so i assume that if the method changed the value of p.Name then it would change the original 'steve' but if I change the value of Name in the (p Person) version I am changing something that will be immediately not accessible since I don't have a handle on the copy?

jocull Over a year ago

Just as a note, there is a performance difference when calling a method by value. It may or may not be significant depending on the size of your structure or how often the methods are called, but it's something to consider :)

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