io.Copy是阻塞的
关了channel也是能读出来的
func main() {
naturals := make(chan int)
squares := make(chan int)
// Counter
go func() {
for x := 0; x < 100; x++ {
naturals <- x
}
close(naturals)
}()
// Squarer
go func() {
for x := range naturals {
squares <- x * x
}
close(squares)
}()
// Printer (in main goroutine)
for x := range squares {
fmt.Println(x)
}
}改进版
func counter(out chan<- int) {
for x := 0; x < 100; x++ {
out <- x
}
close(out)
}
func squarer(out chan<- int, in <-chan int) {
for v := range in {
out <- v * v
}
close(out)
}
func printer(in <-chan int) {
for v := range in {
fmt.Println(v)
}
}
func main() {
naturals := make(chan int)
squares := make(chan int)
go counter(naturals)
go squarer(squares, naturals)
printer(squares)
}这里range filenames，这种方式挺好
// makeThumbnails4 makes thumbnails for the specified files in parallel.
// It returns an error if any step failed.
func makeThumbnails4(filenames []string) error {
errors := make(chan error)
for _, f := range filenames {
go func(f string) {
_, err := thumbnail.ImageFile(f)
errors <- err
}(f)
}
for range filenames {
if err := <-errors; err != nil {
return err // NOTE: incorrect: goroutine leak!
}
}
return nil
}这个程序有一个微妙的bug。当它遇到第一个非nil的error时会直接将error返回到调用方，使得没有一个goroutine去排空errors channel。这样剩下的worker goroutine在向这个channel中发送值时，都会永远地阻塞下去，并且永远都不会退出。这种情况叫做goroutine泄露，可能会导致整个程序卡住或者跑出out of memory的错误。
最简单的解决办法就是用一个具有合适大小的buffered channel，这样这些worker goroutine向channel中发送错误时就不会被阻塞。（一个可选的解决办法是创建一个另外的goroutine，当main goroutine返回第一个错误的同时去排空channel。）
改进版，需要认真体会区别，加了buffer之后不会让goroutine一直阻塞着。
// makeThumbnails5 makes thumbnails for the specified files in parallel.
// It returns the generated file names in an arbitrary order,
// or an error if any step failed.
func makeThumbnails5(filenames []string) (thumbfiles []string, err error) {
type item struct {
thumbfile string
err error
}
ch := make(chan item, len(filenames))
for _, f := range filenames {
go func(f string) {
var it item
it.thumbfile, it.err = thumbnail.ImageFile(f)
ch <- it
}(f)
}
for range filenames {
it := <-ch
if it.err != nil {
return nil, it.err
}
thumbfiles = append(thumbfiles, it.thumbfile)
}
return thumbfiles, nil
}这是另一个版本，更加精妙，特别是closer
// makeThumbnails6 makes thumbnails for each file received from the channel.
// It returns the number of bytes occupied by the files it creates.
func makeThumbnails6(filenames <-chan string) int64 {
sizes := make(chan int64)
var wg sync.WaitGroup // number of working goroutines
for f := range filenames {
wg.Add(1)
// worker
go func(f string) {
defer wg.Done()
thumb, err := thumbnail.ImageFile(f)
if err != nil {
log.Println(err)
return
}
info, _ := os.Stat(thumb) // OK to ignore error
sizes <- info.Size()
}(f)
}
// closer
go func() {
wg.Wait()
close(sizes)
}()
var total int64
for size := range sizes {
total += size
}
return total
}注意Add和Done方法的不对称。Add是为计数器加一，必须在worker goroutine开始之前调用，而不是在goroutine中；否则的话我们没办法确定Add是在"closer" goroutine调用Wait之前被调用。并且Add还有一个参数，但Done却没有任何参数；其实它和Add(-1)是等价的。我们使用defer来确保计数器即使是在出错的情况下依然能够正确地被减掉。上面的程序代码结构是当我们使用并发循环，但又不知道迭代次数时很通常而且很地道的写法。
sizes channel携带了每一个文件的大小到main goroutine，在main goroutine中使用了range loop来计算总和。观察一下我们是怎样创建一个closer goroutine，并让其在所有worker goroutine们结束之后再关闭sizes channel的。两步操作：wait和close，必须是基于sizes的循环的并发。
考虑一下另一种方案：如果等待操作被放在了main goroutine中，在循环之前，这样的话就永远都不会结束了，如果在循环之后，那么又变成了不可达的部分，因为没有任何东西去关闭这个channel，这个循环就永远都不会终止。
加了buffer，这样处理的时候有内容才能处理
// tokens is a counting semaphore used to
// enforce a limit of 20 concurrent requests.
var tokens = make(chan struct{}, 20)
func crawl(url string) []string {
fmt.Println(url)
tokens <- struct{}{} // acquire a token
list, err := links.Extract(url)
<-tokens // release the token
if err != nil {
log.Print(err)
}
return list
}###只打印偶数 很妙很秒，认真体会。这种用法是隔一个搞一个 ch这个channel的buffer大小是1，所以会交替的为空或为满，所以只有一个case可以进行下去，无论i是奇数或者偶数，它都会打印0 2 4 6 8。 但如果是再加buffer就不确定了
ch := make(chan int, 1)
for i := 0; i < 10; i++ {
select {
case x := <-ch:
fmt.Println(x) // "0" "2" "4" "6" "8"
case ch <- i:
}
}###一个经典的用go routine换共享变量的例子
// Package bank provides a concurrency-safe bank with one account.
package bank
var deposits = make(chan int) // send amount to deposit
var balances = make(chan int) // receive balance
func Deposit(amount int) { deposits <- amount }
func Balance() int { return <-balances }
func teller() {
var balance int // balance is confined to teller goroutine
for {
select {
case amount := <-deposits:
balance += amount
case balances <- balance:
}
}
}
func init() {
go teller() // start the monitor goroutine
}var (
entering = make(chan client)
leaving = make(chan client)
messages = make(chan string) // all incoming client messages
)
func broadcaster() {
clients := make(map[client]bool) // all connected clients
for {
select {
case msg := <-messages:
// Broadcast incoming message to all
// clients' outgoing message channels.
for cli := range clients {
cli <- msg
}
case cli := <-entering:
clients[cli] = true
case cli := <-leaving:
delete(clients, cli)
close(cli)
}
}
}var (
sema = make(chan struct{}, 1) // a binary semaphore guarding balance
balance int
)
func Deposit(amount int) {
sema <- struct{}{} // acquire token
balance = balance + amount
<-sema // release token
}
func Balance() int {
sema <- struct{}{} // acquire token
b := balance
<-sema // release token
return b
}对于cpu和编译器来说，可能会打乱代码执行顺序
比如一个map，为啥读的时候要加rlock，是因为rwlock完数据不一定真正刷到内存
只有rlock会强制刷一把，让read操作得到正确的值
func main() {
go func() {
Fun1()
}()
<-ready
}
var ready = make(chan struct{})
func Fun1() {
time.Sleep(time.Second*10)
close(ready)
}type entry struct {
res result
ready chan struct{} // closed when res is ready
}
func New(f Func) *Memo {
return &Memo{f: f, cache: make(map[string]*entry)}
}
type Memo struct {
f Func
mu sync.Mutex // guards cache
cache map[string]*entry
}
func (memo *Memo) Get(key string) (value interface{}, err error) {
memo.mu.Lock()
e := memo.cache[key]
if e == nil {
// This is the first request for this key.
// This goroutine becomes responsible for computing
// the value and broadcasting the ready condition.
e = &entry{ready: make(chan struct{})}
memo.cache[key] = e
memo.mu.Unlock()
e.res.value, e.res.err = memo.f(key)
close(e.ready) // broadcast ready condition
} else {
// This is a repeat request for this key.
memo.mu.Unlock()
<-e.ready // wait for ready condition
}
return e.res.value, e.res.err
}
ready用来进行通知所有等待的人==>这不就是singleflight;)不过singleflight是锁。两步锁 跟PL的9.7章节类似