How to initialize members in Go struct

Question

I am new to Golang so allocation in it makes me insane:

import \"sync\"

type SyncMap struct {
        lock *sync.RWMutex
        hm map[string]string
}
func         


        

Answer 1

You just need a constructor. A common used pattern is

func NewSyncMap() *SyncMap {
    return &SyncMap{hm: make(map[string]string)}
}

In case of more fields inside your struct, starting a goroutine as backend, or registering a finalizer everything could be done in this constructor.

func NewSyncMap() *SyncMap {
    sm := SyncMap{
        hm: make(map[string]string),
        foo: "Bar",
    }

    runtime.SetFinalizer(sm, (*SyncMap).stop)

    go sm.backend()

    return &sm
}

Answer 2

The solution of 'Mue' doesn't work since the mutex is not initialized. The following modification works:

package main

import "sync"

type SyncMap struct {
        lock *sync.RWMutex
        hm map[string]string
}

func NewSyncMap() *SyncMap {
        return &SyncMap{lock: new(sync.RWMutex), hm: make(map[string]string)}
}

func (m *SyncMap) Put (k, v string) {
        m.lock.Lock()
        defer m.lock.Unlock()
        m.hm[k] = v
}

func main() {
    sm := NewSyncMap()
    sm.Put("Test", "Test")
}

http://play.golang.org/p/n-jQKWtEy5

Answer 3

Good catch by deamon. Mue was possibly thinking of the more common pattern of including the lock as a value rather than a pointer. Since the zero value of a Mutex is a ready-to-use unlocked Mutex, it requires no initialization and including one as a value is common. As a further simplification, you can embed it by omitting the field name. Your struct then acquires the method set of the Mutex. See this working example, http://play.golang.org/p/faO9six-Qx. Also I took out the use of defer. To some extent it's a matter of preference and coding style, but since it does have a small overhead, I tend not to use it in small functions, especially if there is no conditional code.