Question

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请参阅“A”附近的代码。有人能解释一下这里发生的事情吗？我认为，由于通道没有缓冲，我保证在将数组的切片发布到通道后，我确保在继续之前消耗结果。

???

如果我不复制数据，我会得到以下结果：

package main

import (
    "fmt"
)

func swap(A []int, i int, j int) {
    t := A[i]
    A[i] = A[j]
    A[j] = t
}

func recurse(A []int, c chan []int, depth int) {

    if depth == len(A) {
        // ??? Why do I need to copy the data?
        // If I do c <- A I get an incorrect answer.
        ra := make([]int, len(A))
        copy(ra, A)
        c <- ra
        return
    }

    for i := depth; i < len(A); i++ {
        swap(A, depth, i)
        recurse(A, c, depth+1)
        swap(A, depth, i)
    }
}

func yieldPermutations(A []int, c chan []int) {
    recurse(A, c, 0)
    close(c)
}

func main() {
    A := []int{1, 2, 3}
    c2 := make(chan []int)

    go yieldPermutations(A, c2)
    for v := range c2 {
        fmt.Println(v)
    }
}

显然，正确的结果（我们通过数据复制获得）是：

[1 3 2]
[1 3 2]
[2 3 1]
[2 3 1]
[3 1 2]
[3 1 2]

Answer 1

认为这段代码就像Python中的生成器/产量一样是错误的，而这是造成错误的原因。

在Python中，当您从生成器请求下一个项目时，生成器开始执行并在到达下一个yield <value>语句时停止。 Python的生成器中没有并行性：消费者运行直到它想要一个值，然后生成器运行直到它产生一个值，然后消费者获取值并继续执行。

在你的go代码中，goroutine与消耗项目的代码同时执行。一旦从主代码中从通道读取项目，goroutine就会同时生成下一个项目。 goroutine和消费者都会一直运行，直到它们到达通道发送/接收，然后将值从goroutine发送到消费者，然后它们都继续执行。

这意味着A的后备数组会被同时修改，因为goroutine会生成下一个项目。这是一种导致意外输出的竞争条件。要证明这是一场比赛，请在频道发送后插入time.Sleep(time.Second)。然后代码产生正确的结果（虽然很慢）：https://play.golang.org/p/uEa_k6Brcc

golang用通道实现发生器/产量：奇数通道行为

1 个答案: