A simple way to ensure interface won't be implemented


Sample code is quite easy:

type Car interface {
    impl() Car

type Animal interface {
    impl() Animal

Now, let's create a structure type:

type Duck struct {

Now, if you add func (d *Duck) Move(), you won't be a car accidently! What if you want to embed two interfaces?

type Duck struct {

The compiler would refuse the code since: Duck.impl is ambiguous, you can't have two methods with the same name in Go definition. So we can use this to create something just like impl Trait for Type in Rust, although this is a workaround,but anyway I work with Go so I have to find out how to ensure this when we need it.

If you want to know what if we want something just take type has Move() method? Then just define:

type Movable interface {

The point is correctly reduce the concept in interface, and do not create interface has general name, unless you know what are you doing,but if you need a workaround, this is.

Since Go can't write: func foo(bar A + B + C), A, B, C are interface, so I suggest write:

func foo(bar interface{

NOTE: newline is required, because Go compiler is stupid. You might want define a private interface for this function only.

Sad thing is we can't use A | B | C, have to do runtime assertion for this.

One more bad thing is because you embedded the interface, Go won't check you provide the implementation of methods or not, you have to do this manually.

Date: 2019-06-18 Tue 00:00

Author: Lîm Tsú-thuàn