英文:
WaitGroup - fatal error: all goroutines are asleep - deadlock
问题
有人能解释一下为什么这段代码会抛出一个"fatal error: all goroutines are asleep - deadlock!"的错误吗?
我找不到问题出在哪里。我看到一些关于这个特定错误的问题,但原因大多是在没有关闭通道的情况下循环遍历通道。
谢谢!
package main
import (
	"fmt"
	"sync"
	"time"
)
func main() {
	ch := make(chan time.Duration)
	var wg sync.WaitGroup
	for _, v := range []time.Duration{5, 1} {
		wg.Add(1)
		go wait(v, ch, wg)
		fmt.Println(<-ch)
	}
	wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration, wg sync.WaitGroup) {
	defer wg.Done()
	time.Sleep(seconds * time.Second)
	c <- seconds
}
这段代码中的问题在于wait函数的参数wg sync.WaitGroup是按值传递的,而不是按引用传递。这意味着在wait函数中对wg的修改不会影响到main函数中的wg。因此,wg.Wait()在等待的时候,wait函数中的wg.Done()没有被调用,导致所有的goroutine都无法退出,最终引发了死锁错误。
要解决这个问题,你可以将wg sync.WaitGroup作为指针传递给wait函数,这样wait函数中对wg的修改就会影响到main函数中的wg。修改后的代码如下:
package main
import (
	"fmt"
	"sync"
	"time"
)
func main() {
	ch := make(chan time.Duration)
	var wg sync.WaitGroup
	for _, v := range []time.Duration{5, 1} {
		wg.Add(1)
		go wait(v, ch, &wg)
		fmt.Println(<-ch)
	}
	wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration, wg *sync.WaitGroup) {
	defer wg.Done()
	time.Sleep(seconds * time.Second)
	c <- seconds
}
这样修改后,代码就不会再抛出死锁错误了。
英文:
Can someone explain why this code throws an "fatal error: all goroutines are asleep - deadlock!"?
I can't seem to find what is wrong. I've seen some questions about this specific error, but the reason was mostly looping through a channel without closing it.
Thank you!
package main
import (
	"fmt"
	"sync"
	"time"
)
func main() {
	ch := make(chan time.Duration)
	var wg sync.WaitGroup
	for _, v := range []time.Duration{5, 1} {
		wg.Add(1)
		go wait(v, ch, wg)
		fmt.Println(<-ch)
	}
	wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration, wg sync.WaitGroup) {
	defer wg.Done()
	time.Sleep(seconds * time.Second)
	c <- seconds
}
答案1
得分: 4
你需要通过引用而不是值传递WaitGroup。否则,Done方法将没有效果。该类型的文档中有如下说明:
> 在首次使用后,WaitGroup不能被复制。
将你的代码修复如下,它将正常工作:
func main() {
    ch := make(chan time.Duration)
    var wg sync.WaitGroup
    for _, v := range []time.Duration{5, 1} {
        wg.Add(1)
        go wait(v, ch, &wg)
        fmt.Println(<-ch)
    }
    wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration, wg *sync.WaitGroup) {
    defer wg.Done()
    time.Sleep(seconds * time.Second)
    c <- seconds
}
此外,还可以将此模式表达如下:
func main() {
    ch := make(chan time.Duration)
    var wg sync.WaitGroup
    for _, v := range []time.Duration{5, 1} {
        wg.Add(1)
        go func() {
            defer wg.Done()
            wait(v, ch)
        }()
        fmt.Println(<-ch)
    }
    wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration) {
    time.Sleep(seconds * time.Second)
    c <- seconds
}
这种情况下的好处是,wait函数不需要意识到任何等待组(它可以是第三方函数),并且不会对传递等待组的方式(值传递还是引用传递)产生混淆。
英文:
You have to pass the WaitGroup by reference, not by value. Otherwise the Done has no effect. The documentation of this type says:
> A WaitGroup must not be copied after first use.
Fix your code to this and it will work:
func main() {
	ch := make(chan time.Duration)
	var wg sync.WaitGroup
	for _, v := range []time.Duration{5, 1} {
		wg.Add(1)
		go wait(v, ch, &wg)
		fmt.Println(<-ch)
	}
	wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration, wg *sync.WaitGroup) {
	defer wg.Done()
	time.Sleep(seconds * time.Second)
	c <- seconds
}
It is also common to express this pattern as follows:
func main() {
	ch := make(chan time.Duration)
	var wg sync.WaitGroup
	for _, v := range []time.Duration{5, 1} {
		wg.Add(1)
		go func() {
			defer wg.Done()
			wait(v, ch)
		}()
		fmt.Println(<-ch)
	}
	wg.Wait()
}
func wait(seconds time.Duration, c chan time.Duration) {
	time.Sleep(seconds * time.Second)
	c <- seconds
}
What's nice about this case is that wait doesn't have to be aware of any wait groups (it could be a 3rd-party function, for example), and there's no confusion about passing a wait group by value or reference.
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