Why does Gob throw 'gob: type not registered for interface: map[string]interface {}' when encoding a map[string]interface{}?

Susan Sarandon
Release: 2024-11-19 13:06:03
Original
935 people have browsed it

Why does Gob throw

gob: Encoding Maps with Interfaces

When attempting to encode a map[string]interface{} using Gob, users may encounter the error message: "gob: type not registered for interface: map[string]interface {}." This error occurs because Gob requires that the type of the data being encoded be registered before it can be processed.

The solution to this problem is straightforward: register the type with Gob using the gob.Register function. In this case, the following code should be added to the program:

gob.Register(map[string]interface{}{})
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This registration step informs Gob that it should be able to encode and decode maps with string keys and interface values.

To demonstrate this, consider the following revised code:

package main

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

func CloneObject(a, b interface{}) []byte {
    gob.Register(map[string]interface{}{})
    buff := new(bytes.Buffer)
    enc := gob.NewEncoder(buff)
    dec := gob.NewDecoder(buff)
    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))
}
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Now, when this code is run, the Gob encoder will successfully encode the map[string]interface{} into a byte array. The error message will no longer be displayed.

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