英文:
Why is the following program deadlock?
问题
以下是您提供的代码的中文翻译：
package main
import "fmt"
func main() {
c := make(chan int)
c <- 5
a := <- c
fmt.Println(a)
}
在这段代码中，数据被发送到通道并从通道接收，但仍然显示错误：
fatal error: all goroutines are asleep - deadlock!
请注意，这段代码中存在死锁错误。死锁通常发生在并发程序中，当所有的goroutine都无法继续执行时发生。在这种情况下，主goroutine在发送数据到通道后被阻塞，而没有其他goroutine来接收数据，导致死锁错误。要解决这个问题，您可以使用goroutine来接收数据或使用带有缓冲区的通道。
英文:
package main
import "fmt"
func main() {
c := make(chan int)
c <- 5
a := <- c
fmt.Println(a)
}
Here both the data is being sent and recieved to/from the channel but still the error is being displayed:
fatal error: all goroutines are asleep - deadlock!
答案1
得分: 2
你的通道 c 没有任何缓冲区（甚至没有一个单元格），这意味着主 goroutine 在 c <- 5 这一行上被锁住了。由于它是你应用程序中唯一的 goroutine，这导致了死锁。
英文:
Your channel c doesn't have any buffer (even a single cell), which means that the main goroutine locks at the c <- 5 line. Since it is the only goroutine in your application, it creates a deadlock.
答案2
得分: 2
该通道是无缓冲的。因此，如果有一个goroutine向其写入数据，并且没有其他goroutine从中读取数据，那么就会发生死锁。
要修复这个问题，可以将其改为无缓冲的通道：
c := make(chan int, 1)
或者从另一个goroutine中异步写入数据：
go func() {
c <- 5
}()
https://play.golang.org/p/RRJILbuZKTQ
由于通道用于在goroutine之间进行通信，很少有情况下会在同一个goroutine中既写入通道又读取通道。一个例外是，如果将通道用作资源池，并且希望预先填充（缓冲）通道。
英文:
The channel is unbuffered. So if one writes to it - and no other goroutine is reading from it - it is a guaranteed deadlock.
To fix, either make it unbuffered:
c := make(chan int, 1)
or write asynchronously from another goroutine:
go func() {
c <- 5
}()
https://play.golang.org/p/RRJILbuZKTQ
Since channels are used to communicate between goroutines, its extremely rare one writes to a channel from the same goroutine as the reads. An exception may be, if using a channel as a resource pool and one wants to pre-populate the (buffered) channel.
答案3
得分: 1
通道 make(chan int) 具有隐式大小为零（参考：https://golang.org/ref/spec#Making_slices_maps_and_channels）
大小为零的通道是无缓冲的。指定大小的通道 make(chan int, n) 是有缓冲的。有关有缓冲和无缓冲通道的讨论，请参阅 http://golang.org/ref/spec#Send_statements。示例 http://play.golang.org/p/VZAiN1V8-P 说明了它们之间的区别。
在这里，channel <- c 或 c <- 的操作将被阻塞，直到有其他协程处理它。因此，将 c := make(chan int) 更改为 c := make(chan int, 1)
package main
import "fmt"
func main() {
c := make(chan int, 1)
c <- 5
a := <- c
fmt.Println(a)
}
英文:
Channels make(chan int) has implicit size zero ( ref: https://golang.org/ref/spec#Making_slices_maps_and_channels)
A channel of size zero is unbuffered. A channel of specified size make(chan int, n) is buffered. See http://golang.org/ref/spec#Send_statements for a discussion on buffered vs. unbuffered channels. The example at http://play.golang.org/p/VZAiN1V8-P illustrates the difference.
Here, channel <-c or c <- will be blocked until someone processes it (concurrently). So, change c := make(chan int) to c := make(chan int,1)
package main
import "fmt"
func main() {
c := make(chan int, 1)
c <- 5
a := <- c
fmt.Println(a)
}
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