问题描述
我正在尝试修复golang中的竞赛条件,我使用了waitgroup,因为您可以看到我添加了两个waitgroup并调用了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()
}
程序输出上方
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
...
解决方法
任何正在寻找如何使用sync.Mutex在golang中进行同步的人。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,&mux)
wg.Wait()
}