我是Go通道的新手,我正在尝试通过构建模拟内核并通过通道处理交互来学习Go通道。此示例程序的目的是让多个进程 (2) 使用单通道同时向内核发送内存分配请求,其他进程使用单个但不同的通道向内核发送释放内存请求。
+-------------+
+------------------+ | |
-> Alloc. Mem. Ch. |<--\ | |
+-----------------+ ---/ +------------------+ >-->| Kernel |
| Process A |<-- +------------------+ -/ | |
+-----------------+ \--> | Realse Mem. Ch. |< | |
+------------------+ +-------------+
如果我只有分配请求,程序就可以工作,一旦我引入释放请求,程序就会陷入死锁。
请注意,进程在发送分配请求时还会创建一个回复队列,但这在上图中未显示,因为它不是问题的一部分。
完整程序如下:
package main
import (
"fmt"
// "log"
"time"
)
const (
_ float64 = iota
LowPrio
MedPrio
HghPrio
)
// Kernel type to communicate between processes and memory resources
type Kernel struct {
reqMemCh chan chan int
rlsMemCh chan int
}
func (k *Kernel) Init() {
k.reqMemCh = make(chan chan int, 2)
k.rlsMemCh = make(chan int, 2)
go k.AllocMem()
go k.RlsMem()
}
// Fetch memory on process request
func (k *Kernel) GetReqMemCh() chan chan int {
return k.reqMemCh
}
func (k *Kernel) GetRlsMemCh() chan int {
return k.rlsMemCh
}
func (k *Kernel) AllocMem() {
// loop over the items (process reply channels) received over
// the request channel
for pCh := range k.GetReqMemCh() {
// for now think 0 is the available index
// send this as a reply to the exclusive process reply channel
pCh <- 0
close(pCh)
}
}
// Release memory
func (k *Kernel) RlsMem() {
// we do not have to anything here
}
// Process type which requests memory
type Proc struct {
ind int
prio float64
exeT time.Time
count int
memInd int
rqMemCh chan chan int
rlMemCh chan int
}
func (p *Proc) Init(
ind int,
prio float64,
rqMemCh chan chan int,
rlMemCh chan int,
) {
p.ind = ind
p.prio = prio
p.memInd = -1
p.rqMemCh = rqMemCh
p.rlMemCh = rlMemCh
}
func (p *Proc) GetReqMemCh() chan chan int {
return p.rqMemCh
}
func (p *Proc) GetRlsMemCh() chan int {
return p.rlMemCh
}
func (p *Proc) ReqMem() {
// create the reply channel exclusive to the process
// this channel will return the allocated memeory id/address
rpCh := make(chan int)
// send the reply channel through the request channel
// to get back the allocation memory id
p.GetReqMemCh() <- rpCh
// Below line is blocking ...
for mi := range rpCh {
p.memInd = mi
}
}
func (p Proc) RlsMem() {
p.GetRlsMemCh() <- 0
}
func (p Proc) String() string {
return fmt.Sprintf(
"Proc(%d): Memory(%d), Count(%d)",
p.ind+1, p.memInd+1, p.count,
)
}
func main() {
k := &Kernel{}
k.Init()
p := &Proc{}
for i := 0; i < 3; i++ {
p.Init(i, LowPrio, k.GetReqMemCh(), k.GetRlsMemCh())
p.ReqMem()
p.RlsMem()
}
time.Sleep(time.Second)
}
例外情况如下:
fatal error: all goroutines are asleep - deadlock!
goroutine 1 [chan send]:
main.Proc.RlsMem(...)
main.go:100
main.main()
main.go:119 +0xc5
goroutine 6 [chan receive]:
main.(*Kernel).AllocMem(0x0?)
main.go:41 +0x5e
created by main.(*Kernel).Init in goroutine 1
main.go:25 +0xc5
exit status 2
任何帮助将不胜感激。
干杯,
DD。
正如 Brits 评论的那样,您有一个缓冲通道已达到其容量,但没有任何内容可供读取。
根据语言之旅 (1 2),发送和接收块,直到另一方准备好。虽然缓冲通道在这里提供了一些宽容,但一旦缓冲区已满,行为是相同的。
这可以通过添加
k.rlsMemCh
的消费者来解决。如果您没有为此计划任何操作,请删除该通道或暂时将其耗尽。
func (k *Kernel) Init() {
k.reqMemCh = make(chan chan int, 2)
k.rlsMemCh = make(chan int, 2)
go k.AllocMem()
go k.RlsMem()
}
func (k *Kernel) AllocMem() {
for pCh := range k.GetReqMemCh() {
pCh <- 0
close(pCh)
}
}
func (k *Kernel) RlsMem() {
// TODO: Add a for-select or for-range over k.rlsMemCh here
}
排水可能看起来像这样:
func (k *Kernel) RlsMem() {
for {
<-k.GetRlsMemCh()
}
}