trying to understand how the Go gob encoder works

Question

in my ambitions to understand how gob work . i have severals question .

i know that gob serialize a go type like struct map or interface(we must register it's real type) but :

func (dec *Decoder) Decode(e interface{}) error
Decode reads the next value from the input stream and stores it in the data represented by the       
empty interface value.
If e is nil, the value will be discarded. 
Otherwise, the value underlying e must be a pointer to the correct type for the next data item received.
If the input is at EOF, Decode returns io.EOF and does not modify e.

i didn't understand nothing in this documentation . what they mean by ( reads the next value from the input stream ) they are one data that we could send it's a struct or a map but not many .what they mean by If e is nil, the value will be discarded. please expert explain to me i'am disasperate all day and ididn't find nothing

Answer 1

Since entering this answer, I learned that OP is trolling us. Stop feeding the troll.

You can write multiple values to a stream. You can read multiple values from a stream.

This code writes two values to output stream w, an io.Writer:

e := gob.NewEncoder(w)
err := e.Encode(v1)
if err != nil {
   // handle error
}
err := e.Encode(v2)
if err != nil {
  // handle error
}

This code reads the values from stream r, an io.Reader. Each call to Decode reads a value that was written by a call to Decode.

d := gob.NewDecoder(r)
var v1 V
err := e.Decode(&v1)
if err != nil {
   // handle error
}
var v2 V
err := e.Decode(&v2)
if err != nil {
  // handle error
}

Writing multiple values to a stream gains efficiency because information about each encoded type is written once to the stream.