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

Question

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

One option might be var factoryFn factory = func(v int) IFoo { return CreateFoo(v) }. Then you could remove the explicit type and go with var factoryFn = func(v int) IFoo { return CreateFoo(v) } — Chris Taylor
– Chris Taylor, Commented Mar 27, 2019 at 0:00
Boy I feel silly. That's a great/easy way to get around the restriction. Thanks! I am a little afraid that I'm misunderstanding something important about the relation between "Struct Pointers" and "Interfaces" though, so I'd still like to know why I can't do that assignment. — Bryan
– Bryan, Commented Mar 27, 2019 at 0:05
I was busy typing a more detailed explanation as an answer when your comment came in. You might want to read up on covariance and contravariance, which would help you understand why the function types are not interchangeable. — Chris Taylor
– Chris Taylor, Commented Mar 27, 2019 at 0:12

Chris Taylor · Accepted Answer · 2019-03-27 00:10:32Z

3

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) }

answered Mar 27, 2019 at 0:10

Chris Taylor

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