英文:
Channel causes panic when called on same scope
问题
第一个代码运行成功而第二个代码失败的原因是什么？第二个代码中，我在demo函数中从通道中接收输入，但仍然导致了死锁。有人能简要解释一下吗？我所看到的是，当我在同一作用域中调用通道时，会出现问题。
英文:
Can someone explain to me why the first code runs but the second one fails?
First One :
func main() {
channel := make(chan int)
go demo(channel)
num := <-channel
println(num)
}
func demo(channel chan int) {
channel <- 2
}
Second One:
func main() {
demo()
}
func demo() {
channel := make(chan int)
channel <- 2
num := <-channel
println(num)
}
In the second one, I am taking the input from the channel in the demo function but it still creates a deadlock can someone explain this in brief.
What I can see is when I am calling the channel in the same scope it's causing an issue.
答案1
得分: 5
你的通道是无缓冲的，因此写入它将会阻塞，因为没有其他goroutine可以从中读取。
进行以下更改，它将会运行：
channel := make(chan int, 1)
英文:
Your channel is unbuffered so writing to it will block since no other goroutines can read from it.
Make this change and it will run:
channel := make(chan int, 1)
答案2
得分: 3
在第一种情况下，写入通道的操作在一个单独的 goroutine 中进行。当主 goroutine 阻塞在通道读取时，仍然有一个 goroutine 可以写入通道，因此不会发生死锁。最终，该 goroutine 会写入通道，程序完成。
在第二种情况下，写入通道被阻塞，并且没有其他正在运行的 goroutine，因此会发生死锁。
英文:
In the first case, write to channel is in a separate goroutine. When the main goroutine blocks reading from the channel, there is still one goroutine that can write to the channel, thus is it not a deadlock. Eventually that goroutine writes to the channel and the program completes.
In the second case, write to channel blocks, and there are no other goroutines running, so it is a deadlock.
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