为什么在直接实现的接口上的此类型声明失败？

Question

I am struggling with Go's Type Assertion mechanism. In the below example the Type Assertion for Qux.(Bar) fails.

Why does a direct implementation of DoBar() at Qux not fullfill the Bar interface?

package main

import (
    "fmt"
)

type Nameable interface {
    Name() string
}

type Foo interface {
    Nameable
    DoFoo() string
}

type Bar interface {
    Nameable
    DoBar() string
}

type bar struct {
    name string
}

func (b bar) Name() string {
    return b.name
}

// Qux embeds bar and is expected to fullfill Nameable interface
type Qux struct {
    bar
}

func (q *Qux) DoBar() string {
    return "DoBar"
}

func Check(subject Nameable) {
    if N, ok := subject.(Nameable); ok {
        fmt.Printf("%s has Nameable
", N.Name())
    } 

    if F, ok := subject.(Foo); ok {
        fmt.Printf("%s has Foo: %s
", F.Name(), F.DoFoo())
    }

    if B, ok := subject.(Bar); ok {
        fmt.Printf("%s has Bar: %s
", B.Name(), B.DoBar())
    }
}

func main() {
    Check(bar{name: "bar"})
    Check(Qux{bar: bar{name: "Qux"}})
}

https://play.golang.org/p/PPkUMUu58JW

Output:

bar has Nameable
Qux has Nameable

Answer 1

Qux.DoBar() has pointer receiver, so only *Qux implements Bar but not Qux. The type Qux and the pointer type to it *Qux are different types with different method sets.

Using a value of type *Qux does implement Bar:

Check(&Qux{bar: bar{name: "*Qux"}})

This outputs (try it on the Go Playground):

*Qux has Nameable
*Qux has Bar: DoBar

Also if you change the receiver of Qux.DoBar() to be non-pointer:

func (q Qux) DoBar() string {
    return "DoBar"
}

Then both Qux and *Qux will implement Bar:

Check(bar{name: "bar"})
Check(Qux{bar: bar{name: "Qux"}})
Check(&Qux{bar: bar{name: "*Qux"}})

Output (try it on the Go Playground):

bar has Nameable
Qux has Nameable
Qux has Bar: DoBar
*Qux has Nameable
*Qux has Bar: DoBar

See related question: X does not implement Y (... method has a pointer receiver)