dounang1974 2016-01-18 16:24
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UnmarshalJSON导致堆栈溢出

This works: http://play.golang.org/p/-Kv3xAguDR.

This results in a stack overflow: http://play.golang.org/p/1-AsHFj51O.

I fail to understand why. What is the correct way use the JSONUnmarshaler interface in this case?

package main

import (
    //"bytes"
    "encoding/json"
    "fmt"
    "strings"
)

type T interface {
    Printer()
}

type A struct{ JA string }

func (t A) Printer() { fmt.Print("A") }

/*
func (t *A) UnmarshalJSON(data []byte) error {
    i := A{}
    dec := json.NewDecoder(bytes.NewReader(data))
    if err := dec.Decode(&i); err != nil {
        return err
    }
    i.Printer()
    *t = i
    return nil
}
*/

var vI []T

func main() {
    vI = []T{&A{}}
    get()
}

func get() {
    dec := json.NewDecoder(strings.NewReader("[{\"JA\":\"OK\"}]"))
    if err := dec.Decode(&vI); err != nil {
        fmt.Print(err)
    }
    for _, v := range vI {
        v.Printer()
    }
}
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1条回答 默认 最新

  • dqxafj6830 2016-01-18 16:35
    关注

    This

    dec.Decode(&i)
    

    will call your UnmarshalJSON, which in turn will call Decode, and so on. If you need to unmarshal your JSON and then do something with it, one neat technique is to declare a local type, unmarshal your data into it, and then convert back to your desired type:

    // Type a has no UnmarshalJSON.
    type a A
    i := a{}
    dec := json.NewDecoder(bytes.NewReader(data))
    if err := dec.Decode(&i); err != nil {
        return err
    }
    // Convert back to A.
    tt := A(i)
    tt.Printer()
    *t = tt
    // ...
    

    Playground: http://play.golang.org/p/HWamV3MbvW.

    The type a has no methods (so no stack overflow), but is convertible to A.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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