Question

通过net/http和相关库中的一些go资料，我发现something使我感到好奇。我正在这里查看版本1.12。

func (p *ReverseProxy) handleUpgradeResponse(rw http.ResponseWriter, req *http.Request, res *http.Response) {
  ...
  hj, ok := rw.(http.Hijacker)
  ...
  conn, brw, err := hj.Hijack()
  ...
}

我已经在更多地方以及标准库之外看到了类似的东西。这是怎么回事在实现特定断言之前，是否隐藏了接口实现的某些方法？为什么我不能只调用Hijack()对象上的rw？

Answer 1

方法似乎是隐藏的，因为该函数采用接口类型。 http.ResponseWriter的接口未定义Hijack()方法。这是在http.Hijacker界面中定义的。具体类型可以实现多个接口。但是，即使将这种具体类型传递到其类型定义为接口的范围中，也无法访问其他方法。因此，在有疑问的示例中，执行类型声明以使Hijack()方法可用。

一些示例（playground）：

package main

import (
    "fmt"
)

type Ship interface {
    Load(containers []string)
    Condition() []string
}

type Sea interface {
    Draft() int
}

// seaShip implements the Sea and Ship interfaces
type seaShip struct {
    containers []string
}

// Load is only part of the Ship interface
func (ss *seaShip) Load(containers []string) {
    ss.containers = append(ss.containers, containers...)
}

// Condition is only part of the Ship interface
func (ss *seaShip) Condition() []string {
    return ss.containers
}

// Draft is only part of the Sea interface
func (ss *seaShip) Draft() int {
    return len(ss.containers)
}

// Pirates is not defined in any interface and therefore can only be called on the concrete type
func (ss *seaShip) Pirates() string {
    return "Help!"
}

// NewShip returns an implementation of the Ship interface
func NewShip() Ship {
    return &seaShip{}
}

func main() {
    ship := NewShip()
    ship.Load([]string{"Beer", "Wine", "Peanuts"})
    fmt.Println(ship.Condition())
    // Won't compile, method is not part of interface!
    // fmt.Println(ship.Draft())

    // Assert to make Draft() available
    sea := ship.(Sea)
    fmt.Println(sea.Draft())

    // Won't compile, methods are not part of interface!
    // fmt.Println(sea.Condition())
    // fmt.Println(sea.Pirates())

    // Assert to the concrete type makes all methods available
    ss := sea.(*seaShip)
    fmt.Println(ss.Condition())
    fmt.Println(ss.Pirates())
}

类型实现中的隐藏方法？

1 个答案: