Why Golang cannot generate json from struct with front lowercase character?

Question

I am trying to print json result from struct I created as following:

type Machine struct {
  m_ip string
  m_type string
  m_serial string
}

and print out

m:= &Machine{ m_ip:"test", m_type:"test", m_serial:"test" }
m_json:= json.Marshal(m)
fmt.Println(m_json)

However, result returned just {}

Secondly,I tried to changed the first letter of words to Uppercase as follow:

type Machine struct{
  MachIp string
  MachType string
  MachSerial string
}

and it works! Why doesn't the word with lowercase character at the front work, anyway?

Possible duplicate of stackoverflow.com/questions/11693865/… — jimt
– jimt, Commented Feb 17, 2014 at 10:04
Is there that kind of concept about exported and imported field? Can you give me a link to that information? That will be a great help! — A-letubby
– A-letubby, Commented Feb 18, 2014 at 2:06
No need, I found it now. Totally, understandable. golang.org/ref/spec#Exported_identifiers — A-letubby
– A-letubby, Commented Feb 18, 2014 at 2:17
Possible duplicate of JSON and dealing with unexported fields — Jonathan Hall
– Jonathan Hall, Commented Jun 20, 2017 at 10:43

James Henstridge · Accepted Answer · 2014-02-17 09:40:00Z

82

Go uses case to determine whether a particular identifier is public or private within the context of your package. In your first example, the fields are not visible to json.Marshal because it is not part of the package containing your code. When you changed the fields to be upper case, they became public so could be exported.

If you need to use lower case identifiers in your JSON output though, you can tag the fields with the desired identifiers. For example:

type Machine struct{
    MachIp     string `json:"m_ip"`
    MachType   string `json:"m_type"`
    MachSerial string `json:"m_serial"`
}

answered Feb 17, 2014 at 9:40

James Henstridge

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