英文:
Is it possible to access channels ch1, ch2 using `select` in Golang?
问题
我正在尝试调试这段代码，但卡在这里。我想要访问ch1和ch2，但发现没有打印任何内容。
package main
import (
"fmt"
)
type degen struct {
i, j string
}
func (x degen) CVIO(ch1, ch2 chan string, quit chan int, m, n string) {
for {
select {
case ch1 <- m:
fmt.Println(x.i)
case ch2 <- n:
fmt.Println("ok")
case <-quit:
fmt.Println("quit")
return
}
}
}
func main() {
ch1 := make(chan string)
ch2 := make(chan string)
quit := make(chan int)
x := degen{"goosebump", "ok"}
go x.CVIO(ch1, ch2, quit, "goosebump", "ok")
}
期望的结果：
它应该打印即将产生的通道数据。
英文:
I was trying to debug this code but am stuck here. I wanted to access ch1, ch2 but found printed nothing.
package main
import (
"fmt"
)
type degen struct {
i, j string
}
func (x degen) CVIO(ch1, ch2 chan string, quit chan int, m, n string) {
for {
select {
case ch1 <- m:
fmt.Println(x.i)
case ch2 <- n:
fmt.Println("ok")
case <-quit:
fmt.Println("quit")
return
}
}
}
func main() {
ch1 := make(chan string)
ch2 := make(chan string)
quit := make(chan int)
x := degen{"goosebump", "ok"}
go x.CVIO(ch1, ch2, quit, "goosebump", "ok")
}
Desired:
It should print the channel data as to be produced.
答案1
得分: 3
你的代码的预期功能并不是很清楚：
main()函数在结束时没有等待 go 协程退出（循环可能根本不会运行）。- 在
select语句中，发送操作无法进行，因为没有接收者（参考 规范 - "如果容量为零或不存在，则通道是无缓冲的，只有在发送方和接收方都准备好时通信才会成功。"）。 - 没有向
quit通道发送任何内容。
我怀疑以下代码（playground）可能会实现你的预期功能。
package main
import (
"fmt"
"sync"
)
type degen struct {
i, j string
}
func (x degen) CVIO(ch1, ch2 chan string, quit chan int, m, n string) {
for {
select {
case ch1 <- m:
fmt.Println(x.i)
case ch2 <- n:
fmt.Println("ok")
case <-quit:
fmt.Println("quit")
return
}
}
}
func main() {
ch1 := make(chan string)
ch2 := make(chan string)
quit := make(chan int)
x := degen{"goosebump", "ok"}
var wg sync.WaitGroup
wg.Add(1)
go func() {
x.CVIO(ch1, ch2, quit, "goosebump", "ok")
wg.Done()
}()
<-ch1 // 从 CH1 接收（允许 go 协程中的 "ch1 <- m" 继续执行）
<-ch2 // 从 CH2 接收（允许 go 协程中的 "ch2 <- n" 继续执行）
quit <- 1
wg.Wait() // 等待 CVIO 结束（由于上述发送操作，它应该会结束）
}
希望对你有帮助！
英文:
Its not really clear what you expect your code to do:
main()ends without waiting for the go routine to exit (its quite possible it the loop will not run at all).- in the
selectthe sends will not proceed because there is no receiver (spec - "if the capacity is zero or absent, the channel is unbuffered and communication succeeds only when both a sender and receiver are ready."). - Nothing is sent to the
quitchannel.
I suspect that the following (playground) might do what you were expecting.
package main
import (
"fmt"
"sync"
)
type degen struct {
i, j string
}
func (x degen) CVIO(ch1, ch2 chan string, quit chan int, m, n string) {
for {
select {
case ch1 <- m:
fmt.Println(x.i)
case ch2 <- n:
fmt.Println("ok")
case <-quit:
fmt.Println("quit")
return
}
}
}
func main() {
ch1 := make(chan string)
ch2 := make(chan string)
quit := make(chan int)
x := degen{"goosebump", "ok"}
var wg sync.WaitGroup
wg.Add(1)
go func() {
x.CVIO(ch1, ch2, quit, "goosebump", "ok")
wg.Done()
}()
<-ch1 // Receive from CH1 (allowing "ch1 <- m" in go routine to proceed)
<-ch2 // Receive from CH2 (allowing "ch2 <- n" in go routine to proceed)
quit <- 1
wg.Wait() // Wait for CVIO to end (which it should do due to above send)
}
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