为什么我的golang频道会引发死锁错误？

Question

package main

import (
    "fmt"
    "sync"
)

func push(c chan int,wg sync.WaitGroup) {
    for i := 0; i < 5; i++ {
        c <- i
    }
    wg.Done()
}

func pull(c chan int,wg sync.WaitGroup) {
    for i := 0; i < 5; i++ {
        result,ok := <- c
        fmt.Println(result,ok)
    }
    wg.Done()
}

func main() {
    var wg sync.WaitGroup
    wg.Add(2)
    c := make(chan int)

    go push(c,wg)
    go pull(c,wg)

    wg.Wait()
}

输出：

localhost:src kuankuan$ go run goroutine.go 
0 true
1 true
2 true
3 true
4 true
throw: all goroutines are asleep - deadlock!

goroutine 1 [semacquire]:
sync.runtime_Semacquire(0x42130100, 0x42130100)
    /usr/local/go/src/pkg/runtime/zsema_amd64.c:146 +0x25
sync.(*WaitGroup).Wait(0x42120420, 0x0)
    /usr/local/go/src/pkg/sync/waitgroup.go:79 +0xf2
main.main()
    /Users/kuankuan/go/src/goroutine.go:31 +0xb9

goroutine 2 [syscall]:
created by runtime.main
    /usr/local/go/src/pkg/runtime/proc.c:221
exit status 2

Answer 1

之所以陷入僵局，是因为结构是通过值而不是通过引用传递的。

将WaitGroup传递给函数时，需要传递 指针 而不是值。否则，将使用WaitGroup的副本。

这是您的工作示例：

package main

import (
    "fmt"
    "sync"
)

func push(c chan int,wg *sync.WaitGroup) {
    for i := 0; i < 5; i++ {
        c <- i
    }
    wg.Done()
}

func pull(c chan int,wg *sync.WaitGroup) {
    for i := 0; i < 5; i++ {
        result,ok := <- c
        fmt.Println(result,ok)
    }
    wg.Done()
}

func main() {
    var wg sync.WaitGroup
    wg.Add(2)
    c := make(chan int)

    go push(c,&wg)
    go pull(c,&wg)

    wg.Wait()
}