如何在Go结构中初始化成员 [英] How to initialize members in Go struct
问题描述
我是Golang的新手,因此它的配置让我疯狂:
$ b
importsync
type SyncMap struct {
lock * sync.RWMutex
hm map [string] string
}
func(m * SyncMap)Put(k,v string){
m.lock.Lock()
推迟m.lock.Unlock()
m.hm [k] = v,true
}
code>
以后,我只是打电话:
sm:= new(SyncMap)
sm.Put(Test,Test)
在这一刻,我得到一个零指针恐慌。
我使用另外一个函数来处理它,并在<$ c之后调用它
$ b
func(m * SyncMap)Init(){
m.hm = make(map [string] string)
m.lock = new(sync.RWMutex)
}
但我想知道,是否有可能摆脱这种样板初始化?
<你只是需要一个构造函数。常用的模式是:
$ p $ func NewSyncMap()* SyncMap {
return& SyncMap {hm:make(map [string] string)}
}
一个goroutine作为后端,或者注册一个终结器,一切都可以在这个构造器中完成。
func NewSyncMap()* SyncMap {
sm:= SyncMap {
hm:make(map [string] string),
foo:Bar,
}
runtime.SetFinalizer(sm ,(* SyncMap).stop)
go sm.backend()
return& sm
}
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 (m *SyncMap) Put (k, v string) {
m.lock.Lock()
defer m.lock.Unlock()
m.hm[k] = v, true
}
and later, I just call:
sm := new(SyncMap)
sm.Put("Test, "Test")
At this moment I get a nil pointer panic.
I've worked around it by using another one function, and calling it right after new()
:
func (m *SyncMap) Init() {
m.hm = make(map[string]string)
m.lock = new(sync.RWMutex)
}
But I wonder, if it's possible to get rid of this boilerplate initializing?
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
}
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