go - 如何在golang中解决种族状况？

Question

我正在尝试修复golang中的比赛条件，我使用了waitgroup，因为您可以看到我添加了两个waitgroup并调用Done()。但是仍然无法保护输出。

package main

import (
    "fmt"
    "time"
    "sync"
)

var counter int = 0

func task(m string, n int,  wg *sync.WaitGroup) {
    defer wg.Done()
    for i := 0 ; i < 100 ; i++ {
        counter = n
        time.Sleep(100000)
        fmt.Println(m, counter)
        counter = 0
    }
}

func main() {
    var wg sync.WaitGroup
    wg.Add(1)
    go task("first", 1, &wg) // i want "first" task should get always 1
    wg.Add(1)
    go task("second", 2, &wg) // // i want "second" task should get always 2
    wg.Wait()
}

以上程序输出

second 1
first 1
first 1
second 1
first 2
second 2
second 2
...

预期产量

second 2
first 1
first 1
second 2
first 1
second 2
second 2
...

Answer 1

任何正在寻找如何使用sync.Mutex进行golang同步的用户。这可能会有所帮助。如果此解决方案不正确，请发表评论或回复。

package main

import (
    "fmt"
    "time"
    "sync"
)

var counter int = 0

func task(m string, n int,  wg *sync.WaitGroup, mux *sync.Mutex) {
    
    defer wg.Done()
    for i := 0 ; i < 100 ; i++ {
        mux.Lock()
        //your shared resource
        counter = n
        time.Sleep(100000)
        fmt.Println(m, counter)
        counter = 0
        //end of your shared resource
        mux.Unlock()
    }
}

func main() {
    var wg sync.WaitGroup
    var mux sync.Mutex
    wg.Add(1)
    go task("first", 1, &wg, &mux)
    wg.Add(1)
    go task("second", 2, &wg, &mux)
    wg.Wait()
}