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!
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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() } }
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