英文:
go channel, seems all right, but it gets deadlock
问题
package main
import "fmt"
import "time"
func main() {
c := make(chan int)
c <- 42 // 向通道 c 写入数据
val := <-c // 从通道 c 读取数据
println(val)
}
我认为 c <- 42 是将 42 写入通道 c，然后 val := <-c 是将通道 c 中的值读取到 val 中。
但为什么会出现死锁呢？
英文:
package main
import "fmt"
import "time"
func main() {
c := make(chan int)
c <- 42 // write to a channel
val := <-c // read from a channel
println(val)
}
I think c <- 42 put 42 to channel c, then val := <-c put value in c to val.
but why does it get deadlock?
答案1
得分: 5
你创建了一个非缓冲通道。因此，语句c <- 42会阻塞，直到其他goroutine尝试从通道接收值。由于没有其他goroutine来执行接收操作，你遇到了死锁。有两种方法可以解决这个问题：
- 在一个不同的goroutine中执行接收操作。
- 给通道添加缓冲区。例如，
c := make(chan int, 1)可以让你在通道上发送一个值而不会阻塞。
英文:
You have created an unbuffered channel. So the statement c <- 42 will block until some other goroutine tries to receive a value from the channel. Since no other goroutine is around to do this, you got a deadlock. There are two ways you could fix this:
- Perform the receive in a different goroutine.
- Add a buffer to the channel. For example,
c := make(chan int, 1)would allow you to send a single value on the channel without blocking.
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