How to mark golang struct as implementing interface?

Question

I have interface:

  type MyInterface interface {   ...   } 

and I want to mark that my struct implements it. I think it is not possible in go, but I want to be certain.

I did the following, but I think it results in an anonymous variable that implements interface. Am I right?

  type MyStruct struct {     ...     MyInterface   } 

Answer 1

In Go, implementing an interface is implicit. There is no need to explicitly mark it as implementing the interface. Though it's a bit different, you can use assignment to test if a type implements an interface and it will produce a compile time error if it does not. It looks like this (example from Go's FAQ page);

type T struct{} var _ I = T{}       // Verify that T implements I. var _ I = (*T)(nil) // Verify that *T implements I. 

To answer your second question, yes that is saying your struct is composed of a type which implements that interface.