英文:
Loop values for one go channel to other
问题
请问有人可以帮我检查一下我的代码，看看为什么会出现死锁的情况吗？
package main
import (
"fmt"
"sync"
)
func main() {
myCh := make(chan int, 10)
wg := &sync.WaitGroup{}
wg.Add(10)
// 只读通道
go func(ch <-chan int, wg *sync.WaitGroup) {
value := <-myCh
fmt.Println(value)
wg.Done()
}(myCh, wg)
// 只写通道
go func(ch chan<- int, wg *sync.WaitGroup) {
for i := 0; i < 10; i++ {
myCh <- i
}
wg.Done()
}(myCh, wg)
wg.Wait()
}
我想从发送通道发送循环值，而其他通道只接收每次迭代的值。
英文:
Could anyone please review my code and can check why I am facing a deadlock?
package main
import (
"fmt"
"sync"
)
func main() {
myCh := make(chan int, 10)
wg := &sync.WaitGroup{}
wg.Add(10)
// READ ONLY
go func(ch <-chan int, wg *sync.WaitGroup) {
value := <-myCh
fmt.Println(value)
wg.Done()
}(myCh, wg)
// send ONLY
go func(ch chan<- int, wg *sync.WaitGroup) {
for i := 0; i < 10; i++ {
myCh <- i
}
wg.Done()
}(myCh, wg)
wg.Wait()
}
I want to send the loop values from SEND CHANNEL and OTHER CHANNEL should only recieve the values of each iteration.
答案1
得分: 2
你的读取器go例程在接收到第一个数据后就退出了。你需要在其中加入一个for循环来持续读取该通道。
其次，在外部循环中，你不需要调用wg.Sync()，因为它也会减少等待组。
package main
import (
"fmt"
"sync"
)
func main() {
myCh := make(chan int, 10)
wg := &sync.WaitGroup{}
wg.Add(10)
// 只读
go func(ch <-chan int, wg *sync.WaitGroup) {
for value := range myCh { // 这个循环将持续读取通道
fmt.Println(value)
wg.Done()
}
}(myCh, wg)
// 只发送
go func(ch chan<- int, wg *sync.WaitGroup) {
for i := 0; i < 10; i++ {
myCh <- i
}
// wg.Done() <- 这里不需要
}(myCh, wg)
wg.Wait()
fmt.Println("DONE")
}
希望对你有帮助！
英文:
Your reader go routine exits after the FIRST received data. You have to include a for loop into it to continously read that channel.
Secondly you don't need to call wg.Sync() in your outer loop, because it also decreases the waitgroup.
package main
import (
"fmt"
"sync"
)
func main() {
myCh := make(chan int, 10)
wg := &sync.WaitGroup{}
wg.Add(10)
// READ ONLY
go func(ch <-chan int, wg *sync.WaitGroup) {
for value := range myCh { // this loop will read the channel continously
fmt.Println(value)
wg.Done()
}
}(myCh, wg)
// send ONLY
go func(ch chan<- int, wg *sync.WaitGroup) {
for i := 0; i < 10; i++ {
myCh <- i
}
// wg.Done() <- This not needed here
}(myCh, wg)
wg.Wait()
fmt.Println("DONE")
}
答案2
得分: 1
你在wg.Add(10)中添加了delta = 10，并且只调用了两次wg.Done()。
你应该要么调用wg.Add(2)，要么调用wg.Done()十次。
现在主函数在wg.Wait()处被锁住，等待另外八次wg.Done()的调用。
英文:
You are adding delta = 10 in wg.Add(10), and call wg.Done() only twice.
You should either call wg.Add(2) or call wg.Done() ten times.
Right now the main function locks at wg.Wait() and waits for eight more wg.Done() calls.
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