UnmarshalJSON results in stack overflow

Question

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()
    }
}

Ainar-G · Accepted Answer · 2016-01-18 16:42:12Z

2

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.

edited Jan 18, 2016 at 16:42

answered Jan 18, 2016 at 16:35

Ainar-G

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