文章目录
- 什么是channel
- channel的实现
- 问题
- 参考文献
什么是channel
我们来看《Go语言编程》中的一段话
channel是Go语言在语言级别提供的goroutine间的通信方式,是一种进程内的通信方式。
通俗点儿解释就是channel可以在两个或者多个goroutine之间传递消息。在Go中,goroutine和channel是并发编程的两大基石,goroutine用来执行并发任务,channel用来在goroutine之间来传递消息。
Do not communicate by sharing memory; instead, share memory by communicating.
不要通过共享内存来通信,而要通过通信来实现共享内存。
推荐一本书《Concurrency in Go》,这本书对golang中的并发做了深入的讲解。
channel的实现
1、引入概念
首先我们来看两个例子来简单看下golang中channel是如何使用的:
package main
import (
"fmt"
)
func goroutineA(a <-chan int) {
for {
select {
case val := <-a:
fmt.Println(val)
}
}
}
func main() {
ch := make(chan int)
go goroutineA(ch)
ch <- 3
ch <- 5
}
很简单的一段程序,初始化了一个非缓冲的channel,然后并发一个协程去接收channel中的数据,然后往channel中连续发送两个值,首先大家先理解一组概念,什么是非缓冲型channel和缓冲型channel?对,其实很简单,make时如果channel空间不为0,就是缓冲型的channel。
ch := make(chan int)//非缓冲型
ch := make(chan int, 1024)//缓冲型
如果我们将go goroutineA(ch)这行代码往下移,会发生什么?对,会报
fatal error: all goroutines are asleep - deadlock!
因为channel没有缓冲,也没有正在等待接收的goroutine,这个概念接下来我会讲到。
另外,我们会看到goroutineA的入参是a <-chan int,代表一个只能用于接收的channel,也就是单向channel。
var a <-chan int//单向channel,只用于接收,也就是从channel读取数据
var a chan<- int//单向channel,只用于发送,也就是往channel写入数据
大家一定要清楚接收和发送的概念
接收代表从channel读取数据
发送代表往channel写入数据
再看一个复杂点儿的例子
package main
import (
"fmt"
"os"
"os/signal"
"syscall"
"time"
)
var exit = make(chan string, 1)
func main() {
go dealSignal()
exited := make(chan struct{}, 1)
go channel1(exited)
count := 0
t := time.Tick(time.Second)
Loop:
for {
select {
case <-t:
count++
fmt.Printf("main run %d\n", count)
case <-exited:
fmt.Println("main exit begin")
break Loop
}
}
fmt.Println("main exit end")
}
func dealSignal() {
c := make(chan os.Signal, 1)
signal.Notify(c, os.Interrupt, syscall.SIGTERM)
go func() {
<-c
exit <- "shutdown"
}()
}
func channel1(exited chan<- struct{}) {
t := time.Tick(time.Second)
count := 0
for {
select {
case <-t:
count++
fmt.Printf("channel1 run %d\n", count)
case <-exit:
fmt.Println("channel1 exit")
close(exited)
return
}
}
}
这个例子首先并发出一个dealsign方法,用来接收关闭信号,如果接收到关闭信号后往exit channel发送一条消息,然后并发运行channel1,channel1中定了一个ticker,正常情况下channel1每秒打印第一个case语句,如果接收到exit的信号,进入第二个case,然后关闭传入的exited channel,那么main中的Loop,接收到exited关闭的信号后,打印“main exit begin”, 然后退出循环,进程成功退出。这个例子演示了channel在goroutine中起到的传递消息的作用。这个例子是为了向大家展示channel在多个goroutine之间进行通信。
2、数据结构
channel为什么会天生具备这种传递消息的特性呢,我们不禁对其底层的数据结构产生兴趣,我们来看下runtime/chan.go文件,有关channel的一切底层操作都在这个文件,我们首先来看下数据结构:
type hchan struct {
qcount uint // total data in the queue;chan中的元素总数
dataqsiz uint // size of the circular queue;底层循环数组的size
buf unsafe.Pointer // points to an array of dataqsiz elements,指向底层循环数组的指针,只针对有缓冲的channel
elemsize uint16 //chan中元素的大小
closed uint32 //chan是否关闭
elemtype *_type // element type;元素类型
sendx uint // send index;已发送元素在循环数组中的索引
recvx uint // receive index;已接收元素在循环数组中的索引
recvq waitq // list of recv waiters,等待接收消息的goroutine队列
sendq waitq // list of send waiters,等待发送消息的goroutine队列
// lock protects all fields in hchan, as well as several
// fields in sudogs blocked on this channel.
//
// Do not change another G's status while holding this lock
// (in particular, do not ready a G), as this can deadlock
// with stack shrinking.
lock mutex
}
type waitq struct {
first *sudog
last *sudog
}
创建一个底层数组容量为5,元素类型为int,那么channel的数据结构如下图所示:
3、创建
首先我们先来了解一下 Channel 在 Go 语言中是如何创建的,Go 语言 Channel 的创建都是由 make 关键字完成的,我们在前面介绍slice和map的创建时都介绍了使用 make 关键字初始化数据结构,那么一个问题,那么Go语言是如何实现通过make方式来创建不同的数据结构的呢?
Golang 中所有形如 make(chan int, 10) 在编译期间会先被转换成 OMAKE 类型的节点,随后的类型检查阶段在发现 make 的第一个参数是 Channel 类型时会将 OMAKE 类型的节点转换成 OMAKECHAN:
func typecheck1(n *Node, top int) (res *Node) {
switch n.Op {
case OMAKE:
// ...
switch t.Etype {
case TCHAN:
l = nil
if i < len(args) {
l = args[i]
i++
l = typecheck(l, ctxExpr)
l = defaultlit(l, types.Types[TINT])
if l.Type == nil {
n.Type = nil
return n
}
if !checkmake(t, "buffer", l) {
n.Type = nil
return n
}
n.Left = l
} else {
n.Left = nodintconst(0)
}
n.Op = OMAKECHAN
}
}
OMAKECHAN 类型的节点最终都会在SSA中间代码生成阶段之前被转换成makechan 或者 makechan64 的函数调用:
func walkexpr(n *Node, init *Nodes) *Node {
switch n.Op {
case OMAKECHAN:
size := n.Left
fnname := "makechan64"
argtype := types.Types[TINT64]
if size.Type.IsKind(TIDEAL) || maxintval[size.Type.Etype].Cmp(maxintval[TUINT]) <= 0 {
fnname = "makechan"
argtype = types.Types[TINT]
}
n = mkcall1(chanfn(fnname, 1, n.Type), n.Type, init, typename(n.Type), conv(size, argtype))
}
}
创建channel的时候,其实底层是调用makechan方法,我们来看下源码:
func makechan(t *chantype, size int) *hchan {
elem := t.elem
// compiler checks this but be safe.
if elem.size >= 1<<16 {
throw("makechan: invalid channel element type")
}
if hchanSize%maxAlign != 0 || elem.align > maxAlign {
throw("makechan: bad alignment")
}
mem, overflow := math.MulUintptr(elem.size, uintptr(size))
if overflow || mem > maxAlloc-hchanSize || size < 0 {
panic(plainError("makechan: size out of range"))
}
// Hchan does not contain pointers interesting for GC when elements stored in buf do not contain pointers.
// buf points into the same allocation, elemtype is persistent.
// SudoG's are referenced from their owning thread so they can't be collected.
// TODO(dvyukov,rlh): Rethink when collector can move allocated objects.
var c *hchan
switch {
case mem == 0:
// Queue or element size is zero.
c = (*hchan)(mallocgc(hchanSize, nil, true))
// Race detector uses this location for synchronization.
c.buf = c.raceaddr()
case elem.ptrdata == 0:
// Elements do not contain pointers.
// Allocate hchan and buf in one call.
c = (*hchan)(mallocgc(hchanSize+mem, nil, true))
c.buf = add(unsafe.Pointer(c), hchanSize)
default:
// Elements contain pointers.
c = new(hchan)
c.buf = mallocgc(mem, elem, true)
}
c.elemsize = uint16(elem.size)
c.elemtype = elem
c.dataqsiz = uint(size)
if debugChan {
print("makechan: chan=", c, "; elemsize=", elem.size, "; elemalg=", elem.alg, "; dataqsiz=", size, "\n")
}
return c
}
可以看出makechan中其实主要的代码就是一个switch,针对不同的情况
1、case mem == 0代表无缓冲型channel,只分配hchan本身结构体大小的内存
2、case elem.ptrdata==0 代表元素类型不含指针,只分配hchan本身结构体大小+元素大小*个数的内存,是连续的内存空间
3、default元素类型包括指针,两次分配内存的操作
然后将buf指向对应的地址,然后是hchan中其他变量的赋值。
4、接收
接下来我们来讲channel的接收和发送,我们使用一段程序来进行讲解
func goroutineA(a <-chan int) {
val := <- a
fmt.Println("G1 received data: ", val)
return
}
func goroutineB(b <-chan int) {
val := <- b
fmt.Println("G2 received data: ", val)
return
}
func main() {
ch := make(chan int)
go goroutineA(ch)
go goroutineB(ch)
ch <- 3
time.Sleep(time.Second)
}
首先创建了一个无缓冲型的channel,然后启动两个goroutine去消费channel的数据,紧接着向channel中发送数据。我们一步一步来分析channel是如何接收和发送数据的,首先来看接收,golang中接收channel数据有两种方式:
i <- ch
i, ok <- ch
这两种不同的方法经过编译器的处理都会变成 ORECV 类型的节点,但是后者会在类型检查阶段被转换成 OAS2RECV 节点,我们可以简单看一下这里转换的路线图:
// entry points for <- c from compiled code
//go:nosplit
func chanrecv1(c *hchan, elem unsafe.Pointer) {
chanrecv(c, elem, true)
}
//go:nosplit
func chanrecv2(c *hchan, elem unsafe.Pointer) (received bool) {
_, received = chanrecv(c, elem, true)
return
}
两者用法不同,chanrecv2可以返回channel是否关闭,但是最终调用方法都是chanrecv,我们来看下源码:
func chanrecv(c *hchan, ep unsafe.Pointer, block bool) (selected, received bool) {
if debugChan {
print("chanrecv: chan=", c, "\n")
}
//##################step1####################
if c == nil {
if !block {
return
}
gopark(nil, nil, waitReasonChanReceiveNilChan, traceEvGoStop, 2)
throw("unreachable")
}
//##################step2####################
if !block && (c.dataqsiz == 0 && c.sendq.first == nil ||
c.dataqsiz > 0 && atomic.Loaduint(&c.qcount) == 0) &&
atomic.Load(&c.closed) == 0 {
return
}
var t0 int64
if blockprofilerate > 0 {
t0 = cputicks()
}
lock(&c.lock)
if c.closed != 0 && c.qcount == 0 {
if raceenabled {
raceacquire(c.raceaddr())
}
unlock(&c.lock)
if ep != nil {
typedmemclr(c.elemtype, ep)
}
return true, false
}
if sg := c.sendq.dequeue(); sg != nil {
// Found a waiting sender. If buffer is size 0, receive value
// directly from sender. Otherwise, receive from head of queue
// and add sender's value to the tail of the queue (both map to
// the same buffer slot because the queue is full).
recv(c, sg, ep, func() { unlock(&c.lock) }, 3)
return true, true
}
if c.qcount > 0 {
// Receive directly from queue
qp := chanbuf(c, c.recvx)
if raceenabled {
raceacquire(qp)
racerelease(qp)
}
if ep != nil {
typedmemmove(c.elemtype, ep, qp)
}
typedmemclr(c.elemtype, qp)
c.recvx++
if c.recvx == c.dataqsiz {
c.recvx = 0
}
c.qcount--
unlock(&c.lock)
return true, true
}
if !block {
unlock(&c.lock)
return false, false
}
// no sender available: block on this channel.
gp := getg()
mysg := acquireSudog()
mysg.releasetime = 0
if t0 != 0 {
mysg.releasetime = -1
}
// No stack splits between assigning elem and enqueuing mysg
// on gp.waiting where copystack can find it.
mysg.elem = ep
mysg.waitlink = nil
gp.waiting = mysg
mysg.g = gp
mysg.isSelect = false
mysg.c = c
gp.param = nil
c.recvq.enqueue(mysg)
goparkunlock(&c.lock, waitReasonChanReceive, traceEvGoBlockRecv, 3)
// someone woke us up
if mysg != gp.waiting {
throw("G waiting list is corrupted")
}
gp.waiting = nil
if mysg.releasetime > 0 {
blockevent(mysg.releasetime-t0, 2)
}
closed := gp.param == nil
gp.param = nil
mysg.c = nil
releaseSudog(mysg)
return true, !closed
}
由于源码较多,我们逐个step进行讲解;
(1)step1
如果channel是nil:如果是非阻塞模式,直接返回(false,false);如果是阻塞模式,调用goprak挂起goroutine,会阻塞下去。
//##################step1####################
if c == nil {
if !block {
return
}
gopark(nil, nil, waitReasonChanReceiveNilChan, traceEvGoStop, 2)
throw("unreachable")
}
(2)step2
快速操作(不用获取锁,快速返回),三组条件全部满足,快速返(false,false)
条件1:首先是在非阻塞模式下
条件2:如果是非缓冲型(datasiz=0)并且等待发送goroutine队列为空(sendq.first=nil,就是没人往channel写数据),或者缓冲型channel(datasiz>0)并且buf中没有数据;
条件3:channel未关闭
//##################step2####################
if !block && (c.dataqsiz == 0 && c.sendq.first == nil ||
c.dataqsiz > 0 && atomic.Loaduint(&c.qcount) == 0) &&
atomic.Load(&c.closed) == 0 {
return
}
(3)step3
首先加锁,如果channel已经关闭,并且buf中没有元素,返回对应类型的0值,但是received为false;两种情况
情形1:非缓冲型,channel已关闭
情形2:缓冲型,channel已关闭,并且buf无元素
//##################step3####################
lock(&c.lock)
if c.closed != 0 && c.qcount == 0 {
if raceenabled {
raceacquire(c.raceaddr())
}
unlock(&c.lock)
if ep != nil {
typedmemclr(c.elemtype, ep)
}
return true, false
}
(4)step4
如果等待发送队列中有元素,证明channel已经满了,两种情形
情形1:非缓冲型,无buf
情形2:缓冲型,buf满了
//##################step4####################
if sg := c.sendq.dequeue(); sg != nil {
recv(c, sg, ep, func() { unlock(&c.lock) }, 3)
return true, true
}
两种情形都正常进入recv方法,我们来看下源码:
func recv(c *hchan, sg *sudog, ep unsafe.Pointer, unlockf func(), skip int) {
//##################step4-1####################
if c.dataqsiz == 0 {
if raceenabled {
racesync(c, sg)
}
if ep != nil {
// copy data from sender
recvDirect(c.elemtype, sg, ep)
}
} else {
//##################step4-2####################
// Queue is full. Take the item at the
// head of the queue. Make the sender enqueue
// its item at the tail of the queue. Since the
// queue is full, those are both the same slot.
qp := chanbuf(c, c.recvx)
if raceenabled {
raceacquire(qp)
racerelease(qp)
raceacquireg(sg.g, qp)
racereleaseg(sg.g, qp)
}
// copy data from queue to receiver
if ep != nil {
typedmemmove(c.elemtype, ep, qp)
}
// copy data from sender to queue
typedmemmove(c.elemtype, qp, sg.elem)
c.recvx++
if c.recvx == c.dataqsiz {
c.recvx = 0
}
c.sendx = c.recvx // c.sendx = (c.sendx+1) % c.dataqsiz
}
sg.elem = nil
gp := sg.g
unlockf()
gp.param = unsafe.Pointer(sg)
if sg.releasetime != 0 {
sg.releasetime = cputicks()
}
goready(gp, skip+1)
}
step4-1:如果是非缓冲型,那么直接从发送者的栈copy到接收者的栈
step4-2:缓冲型的,但是buf已经满了,此时recvx和sendx是重合的,如下图
首先将recvx处的元素0拷贝到接收地址,然后将下一个元素5拷贝到sendx,然后recvx和sendx分别加1。
step4-3:然后唤醒等待发送队列中的goroutine,等待调度器调度。
(5)step5
没有等待发送的队列,并且buf中有元素,直接把接收游标处的数据copy到接收数据的地址,然后改变hchan中元素数据。
if c.qcount > 0 {
// Receive directly from queue
qp := chanbuf(c, c.recvx)
if raceenabled {
raceacquire(qp)
racerelease(qp)
}
if ep != nil {
typedmemmove(c.elemtype, ep, qp)
}
typedmemclr(c.elemtype, qp)
c.recvx++
if c.recvx == c.dataqsiz {
c.recvx = 0
}
c.qcount--
unlock(&c.lock)
return true, true
}
(6)step6
如果是非阻塞,那么直接返回;如果是阻塞的,构造sudog,保存各种值;将sudog保存到channel的recvq中,调用goparkunlock将goroutine挂起
if !block {
unlock(&c.lock)
return false, false
}
// no sender available: block on this channel.
gp := getg()
mysg := acquireSudog()
mysg.releasetime = 0
if t0 != 0 {
mysg.releasetime = -1
}
// No stack splits between assigning elem and enqueuing mysg
// on gp.waiting where copystack can find it.
mysg.elem = ep
mysg.waitlink = nil
gp.waiting = mysg
mysg.g = gp
mysg.isSelect = false
mysg.c = c
gp.param = nil
c.recvq.enqueue(mysg)
goparkunlock(&c.lock, waitReasonChanReceive, traceEvGoBlockRecv, 3)
// someone woke us up
if mysg != gp.waiting {
throw("G waiting list is corrupted")
}
gp.waiting = nil
if mysg.releasetime > 0 {
blockevent(mysg.releasetime-t0, 2)
}
closed := gp.param == nil
gp.param = nil
mysg.c = nil
releaseSudog(mysg)
return true, !closed
我们用本节一开始的例子来讲解下,再贴一遍程序
func goroutineA(a <-chan int) {
val := <- a
fmt.Println("G1 received data: ", val)
return
}
func goroutineB(b <-chan int) {
val := <- b
fmt.Println("G2 received data: ", val)
return
}
func main() {
ch := make(chan int)
go goroutineA(ch)
go goroutineB(ch)
ch <- 3
time.Sleep(time.Second)
}
由于我们创建的channel是无缓冲型的,所以两个goroutine启动的G1和G2会被阻塞,G1和G2被加入到recvq中,状态为waiting,等待被唤醒。此时此刻ch如下图:
问题:当一个channel关闭后,我们是否还能从中读出数据?
package main
import "fmt"
func main() {
ch := make(chan int, 6)
ch <- 1
ch <- 2
close(ch)
a := <-ch
fmt.Println(a)
b := <-ch
fmt.Println(b)
c := <-ch
fmt.Println(c)
}
输出会是什么?
1
2
0
我们可以看出,当一个channel关闭后,我们依然可以从中读出数据,如果chan的buf中有元素,则读出的是chan中buf的数据,如果buf为空,则输出对应元素类型的零值。那么我们来看下如下的一段程序:
package main
import (
"fmt"
"os"
"os/signal"
"syscall"
"time"
)
var exit1 = make(chan struct{}, 1)
func main() {
go dealSignal1()
count := 0
t := time.Tick(time.Second)
for {
select {
case <-t:
count++
fmt.Printf("main run %d\n", count)
case <-exit1:
fmt.Println("main exit begin")
}
}
fmt.Println("main exit over")
}
func dealSignal1() {
c := make(chan os.Signal, 2)
signal.Notify(c, os.Interrupt, syscall.SIGTERM)
go func() {
<-c
close(exit1)
}()
}
上面这段程序会有什么问题?
5、发送
我们继续往下走,G1、G2被挂起后,往channel中发送一个数据3,其实调用的是chansend方法,我们还是逐步的去讲解
(1)step1
如果channel=nil,当前goroutine会被挂起
if c == nil {
if !block {
return false
}
gopark(nil, nil, waitReasonChanSendNilChan, traceEvGoStop, 2)
throw("unreachable")
}
(2)step2
依然是一个不加锁的快速操作,三组条件
条件1:非阻塞
条件2:channel未关闭
条件3:channel是非缓冲型,并且等待接收队列为空;或者缓冲型,并且循环数组已经满了
if !block && c.closed == 0 && ((c.dataqsiz == 0 && c.recvq.first == nil) ||
(c.dataqsiz > 0 && c.qcount == c.dataqsiz)) {
return false
}
(3)step3
加锁,如果channel已经关闭,直接panic
lock(&c.lock)
if c.closed != 0 {
unlock(&c.lock)
panic(plainError("send on closed channel"))
}
(4)step4
如果等待接收队列不为空,说明什么?
情形1:非缓冲型,等待接收队列不为空
情形2:缓冲型,等待接收队列不为空(说明buf为空)
两种情形,都是直接将待发送数据直接copy到接收处
if sg := c.recvq.dequeue(); sg != nil {
// Found a waiting receiver. We pass the value we want to send
// directly to the receiver, bypassing the channel buffer (if any).
send(c, sg, ep, func() { unlock(&c.lock) }, 3)//直接从ep copy到sg
return true
}
func send(c *hchan, sg *sudog, ep unsafe.Pointer, unlockf func(), skip int) {
if raceenabled {
if c.dataqsiz == 0 {
racesync(c, sg)
} else {
// Pretend we go through the buffer, even though
// we copy directly. Note that we need to increment
// the head/tail locations only when raceenabled.
qp := chanbuf(c, c.recvx)
raceacquire(qp)
racerelease(qp)
raceacquireg(sg.g, qp)
racereleaseg(sg.g, qp)
c.recvx++
if c.recvx == c.dataqsiz {
c.recvx = 0
}
c.sendx = c.recvx // c.sendx = (c.sendx+1) % c.dataqsiz
}
}
if sg.elem != nil {
sendDirect(c.elemtype, sg, ep)
sg.elem = nil
}
gp := sg.g
unlockf()
gp.param = unsafe.Pointer(sg)
if sg.releasetime != 0 {
sg.releasetime = cputicks()
}
goready(gp, skip+1)
}
两种情形,都直接从一个用一个goroutine操作另一个goroutine的栈,因此在sendDirect方法中会有一次写屏障
func sendDirect(t *_type, sg *sudog, src unsafe.Pointer) {
// src is on our stack, dst is a slot on another stack.
// Once we read sg.elem out of sg, it will no longer
// be updated if the destination's stack gets copied (shrunk).
// So make sure that no preemption points can happen between read & use.
dst := sg.elem
typeBitsBulkBarrier(t, uintptr(dst), uintptr(src), t.size)
// No need for cgo write barrier checks because dst is always
// Go memory.
memmove(dst, src, t.size)
}
(5)step5
如果等待队列为空,并且缓冲区未满,肯定是缓冲型的channel
if c.qcount < c.dataqsiz {
// Space is available in the channel buffer. Enqueue the element to send.
qp := chanbuf(c, c.sendx)
if raceenabled {
raceacquire(qp)
racerelease(qp)
}
typedmemmove(c.elemtype, qp, ep)
c.sendx++
if c.sendx == c.dataqsiz {
c.sendx = 0
}
c.qcount++
unlock(&c.lock)
return true
}
将元素放在sendx处,然后sendx加1,channel总量加1
(6)step6
如果以上情况都没有命中,说明什么?说明channel已经满了,如果是非阻塞的直接返回,否则需要调用gopack将这个goroutine挂起,等待被唤醒。
if !block {
unlock(&c.lock)
return false
}
// Block on the channel. Some receiver will complete our operation for us.
gp := getg()
mysg := acquireSudog()
mysg.releasetime = 0
if t0 != 0 {
mysg.releasetime = -1
}
// No stack splits between assigning elem and enqueuing mysg
// on gp.waiting where copystack can find it.
mysg.elem = ep
mysg.waitlink = nil
mysg.g = gp
mysg.isSelect = false
mysg.c = c
gp.waiting = mysg
gp.param = nil
c.sendq.enqueue(mysg)
goparkunlock(&c.lock, waitReasonChanSend, traceEvGoBlockSend, 3)
// Ensure the value being sent is kept alive until the
// receiver copies it out. The sudog has a pointer to the
// stack object, but sudogs aren't considered as roots of the
// stack tracer.
KeepAlive(ep)
我们对照程序分析下,在前一个小节G1、G2被挂起来了,等待sender的解救;这时候往ch中发送了一个3,(step4)这时sender发现ch的等待接收队列recvq中有receiver,就会出队一个sudog,然后将元素直接copy到sudog的elem处,然后调用goready将G1唤醒,继续执行G1原来的代码,打印出结果。如下图:
6、关闭
close一个channel会调用closechan方法,比较简单,我们也来看下
(1)step1
如果channel为nil,会直接panic
if c == nil {
panic(plainError("close of nil channel"))
}
(2)step2
加锁,如果channel已经关闭,再次关闭会panic
lock(&c.lock)
if c.closed != 0 {
unlock(&c.lock)
panic(plainError("close of closed channel"))
}
(3)step3
首选将hchan对应close标志置为1,然后声明一个链表;将等待接收队列中的所有sudog加入到链表,并将其elem赋予一个相应类型的0值;
c.closed = 1
var glist gList
// release all readers
for {
sg := c.recvq.dequeue()
if sg == nil {
break
}
if sg.elem != nil {
typedmemclr(c.elemtype, sg.elem)
sg.elem = nil
}
if sg.releasetime != 0 {
sg.releasetime = cputicks()
}
gp := sg.g
gp.param = nil
if raceenabled {
raceacquireg(gp, c.raceaddr())
}
glist.push(gp)
}
(4)step4
将所有等待发送队列的sudog加入链表
// release all writers (they will panic)
for {
sg := c.sendq.dequeue()
if sg == nil {
break
}
sg.elem = nil
if sg.releasetime != 0 {
sg.releasetime = cputicks()
}
gp := sg.g
gp.param = nil
if raceenabled {
raceacquireg(gp, c.raceaddr())
}
glist.push(gp)
}
unlock(&c.lock)
(5)step5
唤醒sudog所有goroutine
for !glist.empty() {
gp := glist.pop()
gp.schedlink = 0
goready(gp, 3)
}
问题
问题1:哪些操作会使channel发生panic?
三种情况
情况1:往一个已经关闭的channel写数据
情况2:关闭一个nil的channel
情况3:关闭一个已经关闭的channel
问题2:channel是并发安全的吗?
是
问题3:当一个channel关闭后,我们是否还能从channel读到数据?
可以,只不过接收的是无效数据
问题4:channel发送和接收元素的本质是什么?
值的拷贝
看一段示例
package main
import (
"fmt"
"time"
)
func print(u <-chan int) {
time.Sleep(2 * time.Second)
fmt.Println("print int", <-u)
}
func main() {
c := make(chan int, 5)
a := 0
c <- a
fmt.Println(a)
// modify g
a = 1
go print(c)
time.Sleep(5 * time.Second)
fmt.Println(a)
}
再看一段复杂一点的
package main
import (
"fmt"
"time"
)
type people struct {
name string
}
var u = people{name: "A"}
func printPeople(u <-chan *people) {
time.Sleep(2 * time.Second)
fmt.Println("printPeople", <-u)
}
func main() {
c := make(chan *people, 5)
var a = &u
c <- a
fmt.Println(a)
// modify g
a = &people{name:"B"}
go printPeople(c)
time.Sleep(5 * time.Second)
fmt.Println(a)
}
输出会是什么
&{A}
printPeople &{A}
&{B}
因为chan中保存的是u的地址的值的拷贝,这个地址未发生改变,虽然调用a = &people{name:“B”}重新赋予了a新的地址,但是channel中的未改变。