package mainimport ( "fmt" "sync")// PUT functionfunc 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 函數中的最后兩個打印語句沒有被打印，我試圖根據鍵將值放入映射中。以及在這種情況下如何關閉 hashMap。
3 回答
九州編程
TA貢獻1785條經驗 獲得超4個贊
您需要創建一個頻道，例如
hashMap[key] = make(chan int)由于您不是從通道讀取，因此需要緩沖通道才能使其工作：
key := "xyz"
hashMap[key] = make(chan int, 5)
試試下面的代碼：
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()
}
輸出：
PUT sent 100
PUT sent 100
PUT sent 100
PUT sent 100
PUT sent 100
縹緲止盈
TA貢獻2041條經驗 獲得超4個贊
我解決問題的方法是：
// 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)
}
在函數中的這行代碼hashMap[key] <- value中put，它不能接受參數中定義的value因為chan int是。nilput (hashMap map[string](chan int)
// 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"])
*/
}
在我的代碼示例中。我將cval緩沖通道 100 初始化為相同大小，hashMap并將 cval 作為put函數中的值傳遞。您只能關閉cval而不是 hashMap 本身。
翻閱古今
TA貢獻1780條經驗 獲得超5個贊
此外，您的代碼可以簡化為此。為什么不必要地傳遞參數！一個額外的修改是我采用不同的值來讓你更清楚地理解這個概念。
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()
}
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