本篇内容介绍了“sync.Once 多次调用一次执行的方法”的有关知识,在实际案例的操作过程中,不少人都会遇到这样的困境,接下来就让小编带领大家学习一下如何处理这些情况吧!希望大家仔细阅读,能够学有所成!
demo
package main
import (
"fmt"
"sync"
)
func main() {
var once sync.Once
onceFunc := func() {
fmt.Println("this func do once")
}
done := make(chan bool)
for i := 0; i < 10; i++ {
go func() {
once.Do(onceFunc)
done <- true
}()
}
for i := 0; i < 10; i++ {
<-done
}
}
output
liqiongtao:test liqiongtao$ go run main.go
this func do once
Once源码
// 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 {
m Mutex
done uint32
}
// 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()) {
if atomic.LoadUint32(&o.done) == 1 {
return
}
// Slow-path.
o.m.Lock()
defer o.m.Unlock()
if o.done == 0 {
defer atomic.StoreUint32(&o.done, 1)
f()
}
}
“sync.Once 多次调用一次执行的方法”的内容就介绍到这里了,感谢大家的阅读。如果想了解更多行业相关的知识可以关注亿速云网站,小编将为大家输出更多高质量的实用文章!
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原文链接:https://my.oschina.net/qiongtaoli/blog/3117080