英文:
How to initialize members in Go struct
问题
我是新手,对Golang中的分配使我发疯:
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
}
然后,我只是调用:
sm := new(SyncMap)
sm.Put("Test", "Test")
此时我会得到一个空指针恐慌。
我通过使用另一个函数来解决这个问题,并在new()
之后立即调用它:
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个样板初始化的方式?
英文:
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?
答案1
得分: 70
你只需要一个构造函数。一个常用的模式是
func NewSyncMap() *SyncMap {
return &SyncMap{hm: make(map[string]string)}
}
如果你的结构体内有更多的字段,可以在这个构造函数中启动一个goroutine作为后台,或者注册一个finalizer来完成所有操作。
func NewSyncMap() *SyncMap {
sm := SyncMap{
hm: make(map[string]string),
foo: "Bar",
}
runtime.SetFinalizer(sm, (*SyncMap).stop)
go sm.backend()
return &sm
}
英文:
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
}
答案2
得分: 11
'Mue'的解决方案不起作用,因为互斥锁没有初始化。以下修改后的代码可以正常工作:
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
英文:
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")
}
答案3
得分: 5
好的,以下是翻译好的内容:
deamon的发现很好。Mue可能在考虑将锁作为值而不是指针的更常见的模式。由于Mutex的零值是一个可立即使用的未锁定的Mutex,它不需要初始化,并且将其作为值包含是常见的。进一步简化,您可以通过省略字段名来嵌入它。然后,您的结构体将获得Mutex的方法集。请参见此工作示例,http://play.golang.org/p/faO9six-Qx。此外,我去掉了对defer的使用。在某种程度上,这是一种偏好和编码风格的问题,但由于它确实有一些小的开销,我倾向于在小函数中不使用它,特别是如果没有条件代码。
英文:
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.
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