CODEWITHSUNDEEP
go
Bryan
Bryan

Reputation: 2128

Function that returns a struct pointer as function that returns an interface

I have a package exporting a function that returns a struct pointer that I can't control. I'd like to wrap that function to provide a dependency seam so that I can do some unit testing. The idea is that I want a fake rather than what the external package is giving me.

An example looks like:

/////////////////////////////////////
// External Package I can't control.
/////////////////////////////////////
package ex
type Foo struct {
    v int
}

func (f *Foo) GetV() int {
    return f.v
}

func CreateFoo(v int) *Foo {
    return &Foo{v: v}
}

////////////////////////////////////
// Local Package I'm testing.
////////////////////////////////////
type IFoo interface {
    GetV() int
}

type factory = func(v int) IFoo
var factoryFn factory = ex.CreateFoo

This results in the error: cannot use CreateFoo (type func(int) *Foo) as type func(int) IFoo in assignment

Since I think that *Foo is an IFoo, why not? What am I missing? Is there a way to achieve what I'm trying to do, giving myself a hook like that?

Playground link to example: https://play.golang.org/p/FDckAQGNRgA

Upvotes: 2

Views: 403

Answers (1)

Chris Taylor
Chris Taylor

Reputation: 53709

The function type returning IFoo is a different type to the function returning *Foo even though Foo meets the requirements for IFoo, Go does not support covariance on the return type.

You would need to wrap the function 

var factoryFn = func(v int) IFoo { return CreateFoo(v) }

Upvotes: 3

Related Questions