我想了解golang频道的运作方式。我读了一本关于go语言的书,找到了以下示例。
package main
import (
"fmt"
)
// Send the sequence 2, 3, 4, ... to returned channel
func generate() chan int {
ch := make(chan int)
go func() {
for i := 2; i <= 100 ; i++ {
ch <- i
}
}()
return ch
}
// Filter out input values divisible by 'prime', send rest to returned channel
func filter(in chan int, prime int) chan int {
out := make(chan int)
go func() {
for {
if i := <-in; i%prime != 0 {
out <- i
}
}
}()
return out
}
func sieve() chan int {
out := make(chan int)
go func() {
ch := generate()
for {
prime := <-ch
ch = filter(ch, prime)
out <- prime
}
}()
return out
}
func main() {
primes := sieve()
for {
fmt.Println(<-primes)
}
}
当我运行该程序时,出现了死锁,但是当我将generate函数更改为
// Send the sequence 2, 3, 4, ... to returned channel
func generate() chan int {
ch := make(chan int)
go func() {
for i := 2; ; i++ {
ch <- i
}
}()
return ch
}
然后,程序将运行无限循环,但不会死锁。当我删除for循环中的条件时,为什么会出现死锁?
阻塞原理是什么意思?
你可以看到它在博客文章“图文并茂的自然通道走 ”
对于无缓冲通道:
(摘自2014年2月由William Kennedy撰写的博客文章“ The Channels in Go ” )
Unbuffered channels have no capacity and therefore require both goroutines to be ready to make any exchange.
When a goroutine attempts to write a resource to an unbuffered channel and there is no goroutine waiting to receive the resource, the channel will lock the goroutine and make it wait.
When a goroutine attempts to read from an unbuffered channel, and there is no goroutine waiting to send a resource, the channel will lock the goroutine and make it wait.
That is what happens in your case with your reader:
func main() {
primes := sieve()
for {
fmt.Println(<-primes)
}
}
since primes
is never closed, main
remains blocked.
It (main
) is in step 3:
在第3步中,右侧的goroutine将手放到通道中或执行read。
该goroutine也锁定在通道中,直到交换完成。
发件人从不打电话close(primes)
。
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