假设一支香烟需要三种成分来制作和吸烟：烟草、纸张和火柴。一张桌子周围有三个吸烟者，每个吸烟者都有三种成分中的一种——一个吸烟者有无限供应的烟草，另一个有纸，第三个有火柴。第四方，拥有无限供应的所有东西，随机选择一个吸烟者，并将香烟所需的用品放在桌子上。被选中的吸烟者吸烟，该过程应无限期重复。

示例代码：

package main

import (
	"fmt"
	"math/rand"
	"sync"
	"time"
)

const (
	paper = iota
	grass
	match
)

var smokeMap = map[int]string{
	paper: "paper",
	grass: "grass",
	match: "match",
}

var names = map[int]string{
	paper: "Sandy",
	grass: "Apple",
	match: "Daisy",
}

type Table struct {
	paper chan int
	grass chan int
	match chan int
}

func arbitrate(t *Table, smokers [3]chan int) {
	for {
		time.Sleep(time.Millisecond * 500)
		next := rand.Intn(3)
		fmt.Printf("Table chooses %s: %s\n", smokeMap[next], names[next])
		switch next {
		case paper:
			t.grass <- 1
			t.match <- 1
		case grass:
			t.paper <- 1
			t.match <- 1
		case match:
			t.grass <- 1
			t.paper <- 1
		}
		for _, smoker := range smokers {
			smoker <- next
		}
		wg.Add(1)
		wg.Wait()
	}
}

func smoker(t *Table, name string, smokes int, signal chan int) {
	var chosen = -1
	for {
		chosen = <-signal // blocks

		if smokes != chosen {
			continue
		}

		fmt.Printf("Table: %d grass: %d match: %d\n", len(t.paper), len(t.grass), len(t.match))
		select {
		case <-t.paper:
		case <-t.grass:
		case <-t.match:
		}
		fmt.Printf("Table: %d grass: %d match: %d\n", len(t.paper), len(t.grass), len(t.match))
		time.Sleep(10 * time.Millisecond)
		select {
		case <-t.paper:
		case <-t.grass:
		case <-t.match:
		}
		fmt.Printf("Table: %d grass: %d match: %d\n", len(t.paper), len(t.grass), len(t.match))
		fmt.Printf("%s smokes a cigarette\n", name)
		time.Sleep(time.Millisecond * 500)
		wg.Done()
		time.Sleep(time.Millisecond * 100)
	}
}

const LIMIT = 1

var wg *sync.WaitGroup

func main() {
	wg = new(sync.WaitGroup)
	table := new(Table)
	table.match = make(chan int, LIMIT)
	table.paper = make(chan int, LIMIT)
	table.grass = make(chan int, LIMIT)
	var signals [3]chan int
	// three smokers
	for i := 0; i < 3; i++ {
		signal := make(chan int, 1)
		signals[i] = signal
		go smoker(table, names[i], i, signal)
	}
	fmt.Printf("%s, %s, %s, sit with \n%s, %s, %s\n\n", names[0], names[1], names[2], smokeMap[0], smokeMap[1], smokeMap[2])
	arbitrate(table, signals)
}

输出：

Sandy, Apple, Daisy, sit with
paper, grass, match

Table chooses match: Daisy
Table: 1 grass: 1 match: 0
Table: 1 grass: 0 match: 0
Table: 0 grass: 0 match: 0
Daisy smokes a cigarette
Table chooses paper: Sandy
Table: 0 grass: 1 match: 1
Table: 0 grass: 1 match: 0
Table: 0 grass: 0 match: 0
Sandy smokes a cigarette
Table chooses match: Daisy
Table: 1 grass: 1 match: 0
Table: 1 grass: 0 match: 0
Table: 0 grass: 0 match: 0
Daisy smokes a cigarette

一个理发店有N张沙发和一张理发椅。没有顾客要理发时，理发师便去睡觉。当一个顾客走进理发店时，如果所有的沙发都已被占用，他便离开理发店；否则，如果理发师正在为其他顾客理发，则该顾客就找一张空沙发坐下等待；如果理发师因无顾客正在睡觉，则由新到的顾客唤醒理发师为其理发。在理发完成后，顾客必须付费，直到理发师收费后才能离开理发店。

示例代码：

package main

import (
	"fmt"
	"sync"
	"time"
)

const (
	sleeping = iota
	checking
	cutting
)

var stateLog = map[int]string{
	0: "Sleeping",
	1: "Checking",
	2: "Cutting",
}
var wg *sync.WaitGroup // Amount of potentional customers

type Barber struct {
	name string
	sync.Mutex
	state    int // Sleeping/Checking/Cutting
	customer *Customer
}

type Customer struct {
	name string
}

func (c *Customer) String() string {
	return fmt.Sprintf("%p", c)[7:]
}

func NewBarber() (b *Barber) {
	return &Barber{
		name:  "Sam",
		state: sleeping,
	}
}

// Barber goroutine
// Checks for customers
// Sleeps - wait for wakers to wake him up
func barber(b *Barber, wr chan *Customer, wakers chan *Customer) {
	for {
		b.Lock()
		defer b.Unlock()
		b.state = checking
		b.customer = nil

		// checking the waiting room
		fmt.Printf("Checking waiting room: %d\n", len(wr))
		time.Sleep(time.Millisecond * 100)
		select {
		case c := <-wr:
			HairCut(c, b)
			b.Unlock()
		default: // Waiting room is empty
			fmt.Printf("Sleeping Barber - %s\n", b.customer)
			b.state = sleeping
			b.customer = nil
			b.Unlock()
			c := <-wakers
			b.Lock()
			fmt.Printf("Woken by %s\n", c)
			HairCut(c, b)
			b.Unlock()
		}
	}
}

func HairCut(c *Customer, b *Barber) {
	b.state = cutting
	b.customer = c
	b.Unlock()
	fmt.Printf("Cutting  %s hair\n", c)
	time.Sleep(time.Millisecond * 100)
	b.Lock()
	wg.Done()
	b.customer = nil
}

// customer goroutine
// just fizzles out if it's full, otherwise the customer
// is passed along to the channel handling it's haircut etc
func customer(c *Customer, b *Barber, wr chan<- *Customer, wakers chan<- *Customer) {
	// arrive
	time.Sleep(time.Millisecond * 50)
	// Check on barber
	b.Lock()
	fmt.Printf("Customer %s checks %s barber | room: %d, w %d - customer: %s\n",
		c, stateLog[b.state], len(wr), len(wakers), b.customer)
	switch b.state {
	case sleeping:
		select {
		case wakers <- c:
		default:
			select {
			case wr <- c:
			default:
				wg.Done()
			}
		}
	case cutting:
		select {
		case wr <- c:
		default: // Full waiting room, leave shop
			wg.Done()
		}
	case checking:
		panic("Customer shouldn't check for the Barber when Barber is Checking the waiting room")
	}
	b.Unlock()
}

func main() {
	b := NewBarber()
	b.name = "Rocky"
	WaitingRoom := make(chan *Customer, 5) // 5 chairs
	Wakers := make(chan *Customer, 1)      // Only one waker at a time
	go barber(b, WaitingRoom, Wakers)

	time.Sleep(time.Millisecond * 100)
	wg = new(sync.WaitGroup)
	n := 10
	wg.Add(10)
	// Spawn customers
	for i := 0; i < n; i++ {
		time.Sleep(time.Millisecond * 50)
		c := new(Customer)
		go customer(c, b, WaitingRoom, Wakers)
	}

	wg.Wait()
	fmt.Println("No more customers for the day")
}

输出：

Checking waiting room: 0
Sleeping Barber - 
Customer 120 checks Sleeping barber | room: 0, w 0 - customer: 
Woken by 120
..............
..............
Checking waiting room: 0
No more customers for the day

该问题描述了两个进程，即生产者和使用者，它们共享一个用作队列的公共固定大小的缓冲区。生产者的工作是生成数据，将其放入缓冲区，然后重新开始。同时，消费者正在消耗数据(即，将其从缓冲区中删除)，一次一个片段。问题在于确保生产者不会尝试将数据添加到缓冲区中(如果数据已满)，并且使用者不会尝试从空缓冲区中删除数据。生产者的解决方案是，如果缓冲区已满，要么进入睡眠状态，要么丢弃数据。下次使用者从缓冲区中删除项目时，它会通知生产者，生产者再次开始填充缓冲区。同样，如果使用者发现缓冲区为空，则可以进入睡眠状态。下次生产者将数据放入缓冲区时，它会唤醒休眠的消费者。

示例代码：

package main

import (
	"flag"
	"fmt"
	"log"
	"os"
	"runtime"
	"runtime/pprof"
)

type Consumer struct {
	msgs *chan int
}

// NewConsumer creates a Consumer
func NewConsumer(msgs *chan int) *Consumer {
	return &Consumer{msgs: msgs}
}

// consume reads the msgs channel
func (c *Consumer) consume() {
	fmt.Println("consume: Started")
	for {
		msg := <-*c.msgs
		fmt.Println("consume: Received:", msg)
	}
}

// Producer definition
type Producer struct {
	msgs *chan int
	done *chan bool
}

// NewProducer creates a Producer
func NewProducer(msgs *chan int, done *chan bool) *Producer {
	return &Producer{msgs: msgs, done: done}
}

// produce creates and sends the message through msgs channel
func (p *Producer) produce(max int) {
	fmt.Println("produce: Started")
	for i := 0; i < max; i++ {
		fmt.Println("produce: Sending ", i)
		*p.msgs <- i
	}
	*p.done <- true // signal when done
	fmt.Println("produce: Done")
}

func main() {
	// profile flags
	cpuprofile := flag.String("cpuprofile", "", "write cpu profile to `file`")
	memprofile := flag.String("memprofile", "", "write memory profile to `file`")

	// get the maximum number of messages from flags
	max := flag.Int("n", 5, "defines the number of messages")

	flag.Parse()

	// utilize the max num of cores available
	runtime.GOMAXPROCS(runtime.NumCPU())

	// CPU Profile
	if *cpuprofile != "" {
		f, err := os.Create(*cpuprofile)
		if err != nil {
			log.Fatal("could not create CPU profile: ", err)
		}
		if err := pprof.StartCPUProfile(f); err != nil {
			log.Fatal("could not start CPU profile: ", err)
		}
		defer pprof.StopCPUProfile()
	}

	var msgs = make(chan int)  // channel to send messages
	var done = make(chan bool) // channel to control when production is done

	// Start a goroutine for Produce.produce
	go NewProducer(&msgs, &done).produce(*max)

	// Start a goroutine for Consumer.consume
	go NewConsumer(&msgs).consume()

	// Finish the program when the production is done
	<-done

	// Memory Profile
	if *memprofile != "" {
		f, err := os.Create(*memprofile)
		if err != nil {
			log.Fatal("could not create memory profile: ", err)
		}
		runtime.GC() // get up-to-date statistics
		if err := pprof.WriteHeapProfile(f); err != nil {
			log.Fatal("could not write memory profile: ", err)
		}
		f.Close()
	}
}

输出：

consume: Started
produce: Started
produce: Sending  0
produce: Sending  1
consume: Received: 0
consume: Received: 1
produce: Sending  2
produce: Sending  3
consume: Received: 2
consume: Received: 3
produce: Sending  4
produce: Done

检查点同步是同步多个任务的问题。考虑一个车间，几个工人组装一些机制的细节。当他们每个人都完成他的工作时，他们会把细节放在一起。没有商店，所以先完成部分的工人必须等待其他人才能开始另一个。将细节放在一起是任务在路径分开之前自我同步的检查点。

示例代码：

package main

import (
	"log"
	"math/rand"
	"sync"
	"time"
)

func worker(part string) {
	log.Println(part, "worker begins part")
	time.Sleep(time.Duration(rand.Int63n(1e6)))
	log.Println(part, "worker completes part")
	wg.Done()
}

var (
	partList    = []string{"A", "B", "C", "D"}
	nAssemblies = 3
	wg          sync.WaitGroup
)

func main() {
	rand.Seed(time.Now().UnixNano())
	for c := 1; c <= nAssemblies; c++ {
		log.Println("begin assembly cycle", c)
		wg.Add(len(partList))
		for _, part := range partList {
			go worker(part)
		}
		wg.Wait()
		log.Println("assemble.  cycle", c, "complete")
	}
}

输出：

2019/07/15 16:10:32 begin assembly cycle 1
2019/07/15 16:10:32 D worker begins part
2019/07/15 16:10:32 A worker begins part
2019/07/15 16:10:32 B worker begins part
........
2019/07/15 16:10:32 D worker completes part
2019/07/15 16:10:32 C worker completes part
2019/07/15 16:10:32 assemble.  cycle 3 complete

哲学家就餐问题可以这样表述，假设有五位哲学家围坐在一张圆形餐桌旁，做以下两件事情之一：吃饭，或者思考。吃东西的时候，他们就停止思考，思考的时候也停止吃东西。餐桌中间有一大碗意大利面，每两个哲学家之间有一只餐叉。因为用一只餐叉很难吃到意大利面，所以假设哲学家必须用两只餐叉吃东西。他们只能使用自己左右手边的那两只餐叉。哲学家就餐问题有时也用米饭和筷子而不是意大利面和餐叉来描述，因为很明显，吃米饭必须用两根筷子。

哲学家从来不交谈，这就很危险，可能产生死锁，每个哲学家都拿着左手的餐叉，永远都在等右边的餐叉(或者相反)。即使没有死锁，也有可能发生资源耗尽。例如，假设规定当哲学家等待另一只餐叉超过五分钟后就放下自己手里的那一只餐叉，并且再等五分钟后进行下一次尝试。这个策略消除了死锁(系统总会进入到下一个状态)，但仍然有可能发生"活锁"。如果五位哲学家在完全相同的时刻进入餐厅，并同时拿起左边的餐叉，那么这些哲学家就会等待五分钟，同时放下手中的餐叉，再等五分钟，又同时拿起这些餐叉。

在实际的计算机问题中，缺乏餐叉可以类比为缺乏共享资源。一种常用的计算机技术是资源加锁，用来保证在某个时刻，资源只能被一个程序或一段代码访问。当一个程序想要使用的资源已经被另一个程序锁定，它就等待资源解锁。当多个程序涉及到加锁的资源时，在某些情况下就有可能发生死锁。例如，某个程序需要访问两个文件，当两个这样的程序各锁了一个文件，那它们都在等待对方解锁另一个文件，而这永远不会发生。

示例代码：main.go

package main

import (
	"hash/fnv"
	"log"
	"math/rand"
	"os"
	"sync"
	"time"
)

// Number of philosophers is simply the length of this list.
var ph = []string{"Mark", "Russell", "Rocky", "Haris", "Root"}

const hunger = 3                // Number of times each philosopher eats
const think = time.Second / 100 // Mean think time
const eat = time.Second / 100   // Mean eat time

var fmt = log.New(os.Stdout, "", 0)

var dining sync.WaitGroup

func diningProblem(phName string, dominantHand, otherHand *sync.Mutex) {
	fmt.Println(phName, "Seated")
	h := fnv.New64a()
	h.Write([]byte(phName))
	rg := rand.New(rand.NewSource(int64(h.Sum64())))
	rSleep := func(t time.Duration) {
		time.Sleep(t/2 + time.Duration(rg.Int63n(int64(t))))
	}
	for h := hunger; h > 0; h-- {
		fmt.Println(phName, "Hungry")
		dominantHand.Lock() // pick up forks
		otherHand.Lock()
		fmt.Println(phName, "Eating")
		rSleep(eat)
		dominantHand.Unlock() // put down forks
		otherHand.Unlock()
		fmt.Println(phName, "Thinking")
		rSleep(think)
	}
	fmt.Println(phName, "Satisfied")
	dining.Done()
	fmt.Println(phName, "Left the table")
}

func main() {
	fmt.Println("Table empty")
	dining.Add(5)
	fork0 := &sync.Mutex{}
	forkLeft := fork0
	for i := 1; i < len(ph); i++ {
		forkRight := &sync.Mutex{}
		go diningProblem(ph[i], forkLeft, forkRight)
		forkLeft = forkRight
	}
	go diningProblem(ph[0], fork0, forkLeft)
	dining.Wait() // wait for philosphers to finish
	fmt.Println("Table empty")
}

输出：

Table empty
Mark seated
Mark Hungry
Mark Eating
..................
..................
Haris Thinking
Haris Satisfied
Haris Left the table
Table empty

下面的程序启动两个Goroutines。这两个Goroutine现在并发运行。我们创建了两个无缓冲的通道，并将它们传递给goroutines作为通道接收到的值的参数。

示例代码：main.go

package main

import (
	"fmt"
)

func main() {
	var intSlice = []int{91, 42, 23, 14, 15, 76, 87, 28, 19, 95}
	chOdd := make(chan int)
	chEven := make(chan int)

	go odd(chOdd)
	go even(chEven)

	for _, value := range intSlice {
		if value%2 != 0 {
			chOdd <- value
		} else {
			chEven <- value
		}
	}

}

func odd(ch <-chan int) {
	for v := range ch {
		fmt.Println("ODD :", v)
	}
}

func even(ch <-chan int) {
	for v := range ch {
		fmt.Println("EVEN:", v)
	}
}

输出：

ODD : 91
ODD : 23
EVEN: 42
EVEN: 14
EVEN: 76
ODD : 15
ODD : 87
ODD : 19
ODD : 95

您在代码中调用go协程，但无法判断协程何时结束，值何时传递到缓冲通道。由于此代码是异步的，因此每当协程完成时，它都会将数据写入通道并在另一端读取。在上面的示例中，您只调用了两个go协程，因此行为是确定的，并且在大多数情况下会以某种方式生成相同的输出，但是当您增加go协程时，输出将不相同，并且顺序将不同，除非您使其同步。

示例代码：main.go

package main

import "fmt"

func sum(a []int, c chan int) {
	sum := 0
	for _, v := range a {
		sum += v
	}
	c <- sum
}

func main() {
	a := []int{17, 12, 18, 9, 24, 42, 64, 12, 68, 82, 57, 32, 9, 2, 12, 82, 52, 34, 92, 36}

	c := make(chan int)
	for i := 0; i < len(a); i = i + 5 {
		go sum(a[i:i+5], c)
	}
	output := make([]int, 5)
	for i := 0; i < 4; i++ {
		output[i] = <-c
	}
	close(c)

	fmt.Println(output)
}

输出：

[296 80 268 112 0]

main函数有两个无缓冲通道ch和quit。在斐波那契函数中，select语句会阻塞，直到其中一个情况准备就绪。

示例代码：main.go

package main

import (
	"fmt"
)

func fibonacci(ch chan int, quit chan bool) {
	x, y := 0, 1
	for {
		select {
		case ch <- x: // write to channel ch
			x, y = y, x+y
		case <-quit:
			fmt.Println("quit")
			return
		}
	}
}

func main() {
	ch := make(chan int)
	quit := make(chan bool)
	n := 10

	go func(n int) {
		for i := 0; i < n; i++ {
			fmt.Println(<-ch) // read from channel ch
		}
		quit <- false
	}(n)

	fibonacci(ch, quit)
}

输出：

0
1
1
2
3
5
8
13
21
34

主要功能有两个功能：generator和receiver。我们创建一个c int通道并从生成器函数返回它。在匿名goroutine中运行的for循环将值写入通道c。

示例代码：main.go

package main

import (
	"fmt"
)

func main() {
	c := generator()
	receiver(c)
}

func receiver(c <-chan int) {
	for v := range c {
		fmt.Println(v)
	}
}

func generator() <-chan int {
	c := make(chan int)

	go func() {
		for i := 0; i < 10; i++ {
			c <- i
		}
		close(c)
	}()

	return c
}

输出：

0
1
2
3
4
5
6
7
8
9

该程序从10个Goroutines开始。我们创建了一个ch字符串通道，并通过同时运行10个goroutines将数据写入该通道。箭头相对于通道的方向指定是发送还是接收数据。指向ch的箭头指定我们要写入通道ch。箭头从ch向外指向指定我们正在从通道ch读取。

示例代码：main.go

package main

import (
	"fmt"
	"strconv"
)

func main() {
	ch := make(chan string)

	for i := 0; i < 10; i++ {
		go func(i int) {
			for j := 0; j < 10; j++ {
				ch <- "Goroutine : " + strconv.Itoa(i)
			}
		}(i)
	}

	for k := 1; k <= 100; k++ {
		fmt.Println(k, <-ch)
	}
}

输出：

1 Goroutine : 9
2 Goroutine : 9
3 Goroutine : 2
4 Goroutine : 0
5 Goroutine : 1
6 Goroutine : 5
7 Goroutine : 3
8 Goroutine : 4
9 Goroutine : 7
10 Goroutine : 6
....................