Golang源码系列三:channel实现原理分析(go channel 指针)

网友投稿 1504 2022-08-19

Golang源码系列三:Channel实现原理分析(go channel 指针)

Golang源码系列三:Channel实现原理分析(go channel 指针)

一、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。

二、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 <- chi, 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:nosplitfunc chanrecv1(c *hchan, elem unsafe.Pointer) { chanrecv(c, elem, true)} //go:nosplitfunc 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.rec-+ 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.rec-+ 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.rec-+ 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?

120

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.rec-+ 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)  }

三、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?

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 mainimport ( "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。

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