使用自定义类型实现接口，而无需导入包

Question

Suppose there's a 3rd party package that makes an interface available:

package giantpackage

type UselessInfo struct {
    wontUse     string
    alsoWontUse string
}

type CoolInterface interface {
    DoSomethingAwesome(ui UselessInfo) (string, error)
}

It seems easy to implement as such:

package main

import giantpackage

type Jedi struct {
    name string
    age  int
}

func (j Jedi) DoSomethingAwesome(ui giantpackage.UselessInfo) (string, error)
    return "Hello world.", nil
}

Assuming:
1) I don't actually use the UselessInfo struct in my DoSomethingAwesome function.
2) The package that I have to import is HUGE.
3) The package that I have to import is no longer maintained, and cannot be modified.

My question:
Is there any way to implement CoolInterface without importing giantpackage?

Answer 1

There is not.

To implement the giantpackage.CoolInterface, your type must have a method:

DoSomethingAwesome(giantpackage.UselessInfo) (string, error)

And to have a method that matches this signature, you have to import giantpackage, else using any other type for the parameter, it won't match the required method.

In your comments you indicated you plan to create a library, and providing an implementation of this interface will be a "nice to have" feature to some of the users of your library.

Then the recommended way is to create a "core" package of your library which does not contain this interface implementation and thus does not depend on giantpackage. Users of your library that do not need this will only import your "core" package and so they will also not depend on giantpackage.

Create another "extension" package of your library which will contain the implementation of this interface (and which may also use your "core" package if needed). Users that do need this can import this package as well.