实现原理(当前代码版本go version go1.11.4 )
1.atomic 原子操作计数器，用于记录此Once对象下的done的值，func()方法只执行一次
2.使用Mutex互斥锁保证多协程操作的安全性
由于代码量比较少贴出源码如下：
// Copyright 2009 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.
package sync
import (
"sync/atomic"
)
// Once is an object that will perform exactly one action.
type Once struct {
// done indicates whether the action has been performed.
// It is first in the struct because it is used in the hot path.
// The hot path is inlined at every call site.
// Placing done first allows more compact instructions on some architectures (amd64/x86),
// and fewer instructions (to calculate offset) on other architectures.
done uint32
m Mutex
}
// Do calls the function f if and only if Do is being called for the
// first time for this instance of Once. In other words, given
// var once Once
// if once.Do(f) is called multiple times, only the first call will invoke f,
// even if f has a different value in each invocation. A new instance of
// Once is required for each function to execute.
//
// Do is intended for initialization that must be run exactly once. Since f
// is niladic, it may be necessary to use a function literal to capture the
// arguments to a function to be invoked by Do:
// config.once.Do(func() { config.init(filename) })
//
// Because no call to Do returns until the one call to f returns, if f causes
// Do to be called, it will deadlock.
//
// If f panics, Do considers it to have returned; future calls of Do return
// without calling f.
//
func (o *Once) Do(f func()) {
// Note: Here is an incorrect implementation of Do:
//
// if atomic.CompareAndSwapUint32(&o.done, 0, 1) {
// f()
// }
//
// Do guarantees that when it returns, f has finished.
// This implementation would not implement that guarantee:
// given two simultaneous calls, the winner of the cas would
// call f, and the second would return immediately, without
// waiting for the first's call to f to complete.
// This is why the slow path falls back to a mutex, and why
// the atomic.StoreUint32 must be delayed until after f returns.
if atomic.LoadUint32(&o.done) == 0 {
// Outlined slow-path to allow inlining of the fast-path.
o.doSlow(f)
}
}
func (o *Once) doSlow(f func()) {
o.m.Lock()
defer o.m.Unlock()
if o.done == 0 {
defer atomic.StoreUint32(&o.done, 1)
f()
}
}
golang中sync包中的sync.Once.Do(f func())用于只执行一次的 func()方法，Mutex互斥锁的实现本章不做探讨
调用的例子：
package main
import (
"fmt"
"sync"
)
type SliceNum []int
func NewSlice() SliceNum {
return make(SliceNum, 0)
}
func (s *SliceNum) Add(elem int) *SliceNum {
*s = append(*s, elem)
fmt.Println("add", elem)
fmt.Println("add SliceNum end", s)
return s
}
func (s *SliceNum) test(elem int) *SliceNum {
fmt.Println("test", elem)
fmt.Println("test", s)
return s
}
func main() {
var once sync.Once
s := NewSlice()
once.Do(func() {
s.Add(16) //方法只执行一次
})
}
在执行Do方法时，方法当做参数传入，此时，Do方法内会判断其中的done 变量的值是否等于0，如果等于0即之前没有执行过此方法会继续执行Once的doSlow(f func())方法，方法内部有互斥锁，先上锁，此处再次判断o.done == 0否，如果等于即没有执行过，才进行传入的方法操作，此后在进行+1原子操作，最后解锁，如果下次再次执行此方法，由于done变量已经不等于0，所以不会在执行传入的方法
