我正在閱讀/工作Go Concurrency Patterns: Pipelines and cancel，但我無法理解停止短部分。我們有以下功能：func sq(in <-chan int) <-chan int {    out := make(chan int)    go func() {        for n := range in {            out <- n * n        }        close(out)    }()    return out}func gen(nums ...int) <-chan int {    out := make(chan int)    go func() {        for _, n := range nums {            out <- n        }        close(out)    }()    return out}func merge(cs ...<-chan int) <-chan int {    var wg sync.WaitGroup    out := make(chan int, 1) // enough space for the unread inputs    // Start an output goroutine for each input channel in cs.  output    // copies values from c to out until c is closed, then calls wg.Done.    output := func(c <-chan int) {        for n := range c {            out <- n        }        wg.Done()    }    wg.Add(len(cs))    for _, c := range cs {        go output(c)    }    // Start a goroutine to close out once all the output goroutines are    // done.  This must start after the wg.Add call.    go func() {        wg.Wait()        close(out)    }()    return out}func main() {    in := gen(2, 3)    // Distribute the sq work across two goroutines that both read from in.    c1 := sq(in)    c2 := sq(in)    // Consume the first value from output.    out := merge(c1, c2)    fmt.Println(<-out) // 4 or 9    return    // Apparently if we had not set the merge out buffer size to 1    // then we would have a hanging go routine. }現(xiàn)在，如果您注意到 line 2in merge，它表示我們chan使用buffer大小為 1的 out ，因為這是未讀輸入的足夠空間。不過，我?guī)缀蹩梢钥隙ǎ覀儜?yīng)該分配chan與buffer大小2.根據(jù)此代碼示例：c := make(chan int, 2) // buffer size 2c <- 1  // succeeds immediatelyc <- 2  // succeeds immediatelyc <- 3  // blocks until another goroutine does <-c and receives 1 由于本節(jié)意味著一個chan的buffer大小3將不會阻止。任何人都可以澄清/幫助我理解嗎？