Go：为“一组”单个结果实施 ManyDecode

Question

我已经实现了一个非常简单的解码方法（gob.Decoder现在使用）——这适用于单个响应——它甚至适用于切片，但我需要实现一个 DecodeMany 方法，它能够解码一组单独的响应（不是一片）。

工作解码方法：

var v MyType

_ = Decode(&v)

...

func Decode(v interface{}) error {

   buf, _ := DoSomething() // func DoSomething() ([]byte, error)

   // error handling omitted for brevity

   return gob.NewDecoder(bytes.NewReader(buf)).Decode(v)

}

我试图为 DecodeMany 方法做的是处理不一定是切片的响应：

var vv []MyType

_ = DecodeMany(&vv)

...

func DecodeMany(vv []interface{}) error {

   for _, g := range DoSomething() { // func DoSomething() []struct{Buf []bytes}

      

      // Use g.Buf as an individual "interface{}"

      // want something like:

      var v interface{} /* Somehow create instance of single vv type? */

      _ = gob.NewDecoder(bytes.NewReader(g.Buf)).Decode(v)

      vv = append(vv, v)

   }

   return

}

除了不编译上面还有错误：

不能在 DecodeMany 的参数中使用 &vv（类型 *[]MyType 的值）作为类型 []interface{}

Answer 1

如果你想修改传递的切片，它必须是一个指针，否则你必须返回一个新的切片。此外，如果函数被声明为具有 type 的参数[]interface{}，则只能传递 type 的值而不能传递[]interface{}其他切片类型......除非你使用泛型......这是开始使用 Go 1.18 中引入的泛型的完美示例。更改DecodeMany()为通用的，T类型参数是切片元素类型：拿指针时func DecodeMany[T any](vv *[]T) error {&nbsp; &nbsp; for _, g := range DoSomething() {&nbsp; &nbsp; &nbsp; &nbsp; var v T&nbsp; &nbsp; &nbsp; &nbsp; if err := gob.NewDecoder(bytes.NewReader(g.Buf)).Decode(&v); err != nil {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return err&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; *vv = append(*vv, v)&nbsp; &nbsp; }&nbsp; &nbsp; return nil}这是一个简单的应用程序来测试它：type MyType struct {&nbsp; &nbsp; S int64}func main() {&nbsp; &nbsp; var vv []MyType&nbsp; &nbsp; if err := DecodeMany(&vv); err != nil {&nbsp; &nbsp; &nbsp; &nbsp; panic(err)&nbsp; &nbsp; }&nbsp; &nbsp; fmt.Println(vv)}func DoSomething() (result []struct{ Buf []byte }) {&nbsp; &nbsp; for i := 3; i < 6; i++ {&nbsp; &nbsp; &nbsp; &nbsp; buf := &bytes.Buffer{}&nbsp; &nbsp; &nbsp; &nbsp; v := MyType{S: int64(i)}&nbsp; &nbsp; &nbsp; &nbsp; if err := gob.NewEncoder(buf).Encode(v); err != nil {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; panic(err)&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; result = append(result, struct{ Buf []byte }{buf.Bytes()})&nbsp; &nbsp; }&nbsp; &nbsp; return}此输出（在Go Playground上尝试）：[{3} {4} {5}]返回切片时如果选择返回切片，则不必传递任何内容，但需要对结果进行赋值：func DecodeMany[T any]() ([]T, error) {&nbsp; &nbsp; var result []T&nbsp; &nbsp; for _, g := range DoSomething() {&nbsp; &nbsp; &nbsp; &nbsp; var v T&nbsp; &nbsp; &nbsp; &nbsp; if err := gob.NewDecoder(bytes.NewReader(g.Buf)).Decode(&v); err != nil {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return result, err&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; result = append(result, v)&nbsp; &nbsp; }&nbsp; &nbsp; return result, nil}使用它：vv, err := DecodeMany[MyType]()if err != nil {&nbsp; &nbsp; panic(err)}fmt.Println(vv)在Go Playground试试这个。