发送通道后 Goroutine 未执行

Question

package main

import (
    "fmt"
    "sync"
)

// PUT function
func put(hashMap map[string](chan int), key string, value int, wg *sync.WaitGroup) {
    defer wg.Done()
    fmt.Printf("this is getting printed")
    hashMap[key] <- value
    fmt.Printf("this is not getting printed")
    fmt.Printf("PUT sent %d\n", value)
}

func main() {
    var value int
    var key string
    wg := &sync.WaitGroup{}
    hashMap := make(map[string](chan int), 100)
    key = "xyz"
    value = 100
    for i := 0; i < 5; i++ {
        wg.Add(1)
        go put(hashMap, key, value, wg)
    }
    wg.Wait()
}

put 函数中的最后两个打印语句没有被打印出来，我正在尝试根据键将值放入映射中。

以及在这种情况下如何关闭 hashMap。

Answer 1

1. 您需要创建一个频道，例如hashMap[key] = make(chan int)
2. 由于您不是从通道读取数据，因此您需要缓冲通道使其工作：

    key := "xyz"
    hashMap[key] = make(chan int, 5)

试试下面的代码：

func put(hashMap map[string](chan int), key string, value int, wg *sync.WaitGroup) {
    hashMap[key] <- value
    fmt.Printf("PUT sent %d\n", value)
    wg.Done()
}
func main() {
    var wg sync.WaitGroup
    hashMap := map[string]chan int{}
    key := "xyz"
    hashMap[key] = make(chan int, 5)
    for i := 0; i < 5; i++ {
        wg.Add(1)
        go put(hashMap, key, 100, &wg)
    }
    wg.Wait()
}

输出：

PUT sent 100
PUT sent 100
PUT sent 100
PUT sent 100
PUT sent 100

Answer 2

我的解决方法是：

// PUT function
func put(hashMap map[string](chan int), key string, value int, wg *sync.WaitGroup) {
    defer wg.Done()
    fmt.Printf("this is getting printed")
    hashMap[key] <- value // <-- nil problem
    fmt.Printf("this is not getting printed")
    fmt.Printf("PUT sent %d\n", value)
}

在put函数中的hashMap[key] <- value这行代码，它不能接受value，因为chan int是nil，它是在put(hashMap map[string](chan int)参数中定义的。

// PUT function
func put(hashMap map[string](chan int), cval chan int, key string, value int, wg *sync.WaitGroup) {
    defer wg.Done()
    fmt.Println("this is getting printed")
    cval <- value // put the value in chan int (cval) which is initialized
    hashMap[key] = cval // set the cval(chan int) to hashMap with key
    fmt.Println("this is not getting printed")
    fmt.Printf("PUT sent %s %d\n", key, value)
}

func main() {
    var value int
    wg := &sync.WaitGroup{}

    cval := make(chan int,100)
    hashMap := make(map[string](chan int), 100)

    value = 100
    for i := 0; i < 5; i++ {
        wg.Add(1)
        go put(hashMap, cval, fmt.Sprintf("key%d",i), value, wg)
    }
    wg.Wait()

    /* uncomment to test cval 
    close(cval)

    fmt.Println("Result:",<-hashMap["key2"])
    fmt.Println("Result:",<-hashMap["key1"])
    cval <- 88 // cannot send value to a close channel
    hashMap["key34"] = cval
    fmt.Println("Result:",<-hashMap["key1"])
    */


}

在我的代码示例中。我将cval 缓冲通道100 的大小相同初始化为hashMap，并将cval 作为put 函数中的值传递。您只能关闭 cval 而不能关闭 hashMap 本身。

Answer 3

此外，您的代码可以简化为这样。为什么不必要地传递参数！一个额外的修改是我采用不同的值来让你更清楚地理解这个概念。

package main

import (
    "log"
    "sync"
)

func put(hash chan int, wg *sync.WaitGroup) {
    defer wg.Done()
    log.Println("Put sent: ", <-hash)
}

func main() {
    hashMap := map[string]chan int{}
    key := "xyz"
    var wg sync.WaitGroup
    hashMap[key] = make(chan int, 5)
    for i := 0; i < 5; i++ {
        value := i
        wg.Add(1)
        go func(val int) {
            hashMap[key] <- val
            put(hashMap[key], &wg)
        }(value)
    }
    wg.Wait()
}