リフレクションする方法スライスとリフレクションを新しく追加します.Appendソーススライスにスライスします

Question

reflect.New[]interface{}のように[]intとし、別のスライスに追加します。

私のコードにはエラーがありますが、正しく設定する方法と、reflect.Newとreflect.AppendSliceの使い方を深く理解する方法はわかりません。

package main 

import (
    "fmt" 
    "reflect" 
) 

func main() { 
    var a []int 
    var value reflect.Value = reflect.ValueOf(&a) 


    if !value.CanSet() { 
     fmt.Println("go here", value.CanSet()) 
     value = value.Elem() 
     fmt.Println("go here", value.CanSet()) 
    } 
    fmt.Println("go here", value.Type())  
    s := reflect.New(value.Type()) 
    fmt.Println("go here", s.Elem()) 
    value = reflect.AppendSlice(value, reflect.ValueOf(s.Elem())) 
    value = reflect.AppendSlice(value, reflect.ValueOf([]int{1, 2}))     
    value = reflect.AppendSlice(value, reflect.ValueOf([]int{3, 4, 5, 6, 7, 8, 9})) 

    fmt.Println(value.Kind(), value.Slice(0, value.Len()).Interface()) 
    //>>slice [1 2 3 4 5 6 7 8 9] 
} 

しかし、それはエラーを与える：

panic: reflect: call of reflect.AppendSlice on struct Value 

goroutine 1 [running]: 
panic(0x100a60, 0x1040e1a0) 
    /usr/local/go/src/runtime/panic.go:500 +0x720 
reflect.flag.mustBe(0x99, 0x17) 
    /usr/local/go/src/reflect/value.go:201 +0xe0 
reflect.AppendSlice(0xfa7e0, 0x1040e130, 0x197, 0x1116a0, 0x1040e190, 0x99, 0x0, 0x0, 0x0, 0x2) 
    /usr/local/go/src/reflect/value.go:1825 +0x60 
main.main() 
    /tmp/sandbox476529744/main.go:21 +0x800 

Answer 1

パニックは、この行にある：

value = reflect.AppendSlice(value, reflect.ValueOf(s.Elem())) 

値s.Elem()がreflect.Valueです。

value = reflect.AppendSlice(value, s.Elem()) 

ない根本的な[]intため、reflect.Valueためreflect.Valueを返しreflect.ValueOf(s.Elem())の式：AppendSliceへの呼び出しに直接この値を使用します。ライン21で

playground example

Answer 2

エラー：

value = reflect.AppendSlice(value, reflect.ValueOf(s.Elem())) 

参照ドキュメント：

func AppendSlice(s, t Value) Value 
The slices s and t must have the same element type. 

あなたは値をダンプするとき、あなたは以下を参照してください

Var dump s.Elem(): []int(nil) 
Var dump reflect.ValueOf(s.Elem()): {typ:0xfa840 ptr:0x1040e160 flag:407} 
Var dump value: []int(nil) 

だから、あなただけの必要はs.Elem()
参照：https://play.golang.org/p/KwXRxGyswg

Answer 3

は、この作業のサンプルを試してみてください（The Go Playground）：

package main 

import (
    "fmt" 
    "reflect" 
) 

func main() { 
    s := reflect.New(reflect.TypeOf([]interface{}{})).Elem() 
    s = reflect.Append(s, reflect.ValueOf(1)) 
    s = reflect.AppendSlice(s, reflect.ValueOf([]interface{}{2, 3, 4, 5, 6, 7, 8, 9})) 
    fmt.Println(s) 
} 

出力：

[1 2 3 4 5 6 7 8 9] 

そして、次を参照してください。https://github.com/golang/go/wiki/InterfaceSlice：

The question then, "Why can't I assign any slice to an []interface{}, when I can assign any type to an interface{}?"

Why?

There are two main reasons for this.

The first is that a variable with type []interface{} is not an interface! It is a slice whose element type happens to be interface{}. But even given this, one might say that the meaning is clear.

Well, is it? A variable with type []interface{} has a specific memory layout, known at compile time.

Each interface{} takes up two words (one word for the type of what is contained, the other word for either the contained data or a pointer to it). As a consequence, a slice with length N and with type []interface{} is backed by a chunk of data that is N*2 words long.

This is different than the chunk of data backing a slice with type []MyType and the same length. Its chunk of data will be N*sizeof(MyType) words long.

The result is that you cannot quickly assign something of type []MyType to something of type []interface{}; the data behind them just look different.