waitgroup.Wait() 导致死锁

小伙伴们有没有觉得学习golang很有意思？有意思就对了！今天就给大家带来《waitgroup.Wait() 导致死锁》，以下内容将会涉及到，若是在学习中对其中部分知识点有疑问，或许看了本文就能帮到你！

我试图找出为什么我与 waitgroup.wait() 发生死锁

package main

import (
    "fmt"
    "sync"
)

var wg sync.WaitGroup

func foo(c chan int, i int) {
    defer wg.Done()
    c <- i
}

func main() {
    ch := make(chan int)
    for i := 0; i < 10; i++ {
        wg.Add(1)
        Go foo(ch, i)
    }
    wg.Wait()
    close(ch)
    for item := range ch {
        fmt.Println(item)
    }
}

当我像这样运行它时，它会打印 fatal error: all goroutine are sleep - deadlock!

我尝试将 ch 更改为缓冲通道，这解决了问题。但我很想知道为什么会出现死锁。

正确答案

我已经注释掉了你的程序逻辑不正确的部分：

package main

import (
    "fmt"
    "sync"
)

var wg sync.waitgroup

func foo(c chan int, i int) {
    defer wg.done()
    c <- i
}

func main() {
    ch := make(chan int) // unbuffered channel

    for i := 0; i < 10; i++ {
        wg.add(1)
        go foo(ch, i)
    }

    // wg.wait is waiting for all goroutines to finish but that's
    // only possible if the send to channel succeeds. in this case,
    // it is not possible as your receiver "for item := range ch" is below
    // this. hence, a deadlock.
    wg.wait()

    // ideally, it should be the sender's duty to close the channel.
    // and closing a channel before the receiver where the channel
    // is unbuffered is not correct.
    close(ch)

    for item := range ch {
        fmt.println(item)
    }
}

更正的程序：

package main

import (
    "fmt"
    "sync"
)

var wg sync.WaitGroup

func foo(c chan int, i int) {
    defer wg.Done()
    c <- i
}

func main() {
    ch := make(chan int)

    go func() {
        for item := range ch {
            fmt.Println(item)
        }
    }()

    for i := 0; i < 10; i++ {
        wg.Add(1)
        go foo(ch, i)
    }

    wg.Wait()
    close(ch)
}

终于介绍完啦！小伙伴们，这篇关于《waitgroup.Wait() 导致死锁》的介绍应该让你收获多多了吧！欢迎大家收藏或分享给更多需要学习的朋友吧~编程网公众号也会发布Golang相关知识，快来关注吧！