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Why do I get the 'gob: type not registered for interface: map[string]interface {}' error in Go?

Patricia Arquette
Patricia ArquetteOriginal
2024-11-11 09:29:03400browse

Why do I get the

Error: "gob: type not registered for interface: map[string]interface {}"

In Go, the gob package provides support for encoding and decoding values into a binary representation. However, when trying to encode a map[string]interface{} value using gob, you may encounter the error:

gob: type not registered for interface: map[string]interface {}

This error signifies that the gob package is not aware of the type map[string]interface{}. To resolve this issue, you must explicitly register the type using the gob.Register function.

gob.Register(map[string]interface{}{})

By registering the type, you are informing the gob package that it should recognize and handle map[string]interface{} values during encoding and decoding operations.

Here's a modified version of the code you provided that includes the necessary registration:

package main

import (
    "bytes"
    "encoding/gob"
    "encoding/json"
    "fmt"
    "log"
)

func CloneObject(a, b interface{}) []byte {
    buff := new(bytes.Buffer)
    enc := gob.NewEncoder(buff)
    dec := gob.NewDecoder(buff)
    // Register the map[string]interface{} type for encoding
    gob.Register(map[string]interface{}{})
    err := enc.Encode(a)
    if err != nil {
        log.Panic("e1: ", err)
    }
    b1 := buff.Bytes()
    err = dec.Decode(b)
    if err != nil {
        log.Panic("e2: ", err)
    }
    return b1
}

func main() {
    var a interface{}
    a = map[string]interface{}{"X": 1}
    b2, err := json.Marshal(&a)
    fmt.Println(string(b2), err)

    var b interface{}
    b1 := CloneObject(&a, &b)
    fmt.Println(string(b1))
}

After registering the type, you can successfully encode and decode the map[string]interface{} value using gob.

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