Define method on multiple structs that implement the same interface

Question

Take the following piece of code:

type A struct { … }
func (a *A) Attr() int { … }

type B struct { … }
func (b *B) Attr() int { … }

type I interface{
  Attr() int
}

func (m that implements I) Process() int {
  do something with m.Attr()
}

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

Methods cannot be defined on interfaces so m cannot be of type I. I tried using anonymous fields on A and B but Attr is specific to the associated structs so it can't be implemented on an anonymous field.

I want to avoid copy/pasting the Process() method on A and B since it is exactly the same. I could simply define

func Process(m I) int { … }

instead but it's not very elegant.

How would you go about doing this the go way?

A function that takes an interface is the idiomatic way to do it. — Mr_Pink
– Mr_Pink, Commented Dec 12, 2014 at 21:35
Everything doesn't have to be a method like in some other OOP languages (e.g. Ruby). — Mr_Pink
– Mr_Pink, Commented Dec 12, 2014 at 21:36
I agree that func Process(m I) int { … } seems to be the way to go. play.golang.org/p/t59t5KHlw6 . — Intermernet
– Intermernet, Commented Dec 12, 2014 at 23:38

James Henstridge · Accepted Answer · 2014-12-15 10:53:46Z

2

This is not currently possible in Go. While you can introduce common methods to a number of types via embedding another type within a struct, those methods have no knowledge of the type they have been embedded in.

The usual Go idiom for this pattern is to use a function. As an example, sort package from the standard library. It defines an interface consisting of the methods needed to implement a sort algorithm on a container (Len, Less, and Swap). However, the actual sort algorithm is exposed as a function that takes an argument implementing the interface.

answered Dec 15, 2014 at 10:53

James Henstridge

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