dongyong3223 2019-03-26 23:49
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返回结构指针的函数作为返回接口的函数

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

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1条回答 默认 最新

  • doze79040 2019-03-27 00:10
    关注

    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) }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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