使用切片类型输入和输出通道并行工作例程

Question

我对围棋语言比较陌生。尽管我不希望如此，但我可能会用一个愚蠢的问题来打扰你。我先道歉，以防万一...

下面是我的示例：我定义了一个worker（）函数，它作为一组并发Go例程从main（）调用。输入和输出数据通过片类型[]int的输入和输出通道提供。在一种情况下，一切正常，在另一种情况下，结果有误。请参阅代码中的注释和代码下面的程序输出。

老实说，我看不出这两种代码变体之间的实际区别。我错过了什么？谢谢你的建议！

package main

import "fmt"
import "runtime"

func worker(x_ch <-chan []int, y_ch chan<- []int, wid int) {

    for x := range x_ch {
        y := x
        fmt.Println("   worker", wid, "x:", x)
        fmt.Println("   worker", wid, "y:", y)
        y_ch <- y
    }
}

func main() {

    n_workers := runtime.NumCPU()
    n_len := 4
    n_jobs := 4
    x := make([]int, n_len)
    x_ch := make(chan []int, 10)
    y_ch := make(chan []int, 10)

    for j := 0; j < n_workers; j++ { go worker(x_ch, y_ch, j) }

    for k := 0; k < n_jobs; k++ {

//      variant 1: works!
        x = []int{k, k, k, k}

//      variant 2: doesn't work!
//      for i := range x { x[i] = k }

        fmt.Println("main x:", k, x)
        x_ch <- x
    }

    close(x_ch)

    for i := 0; i < n_jobs; i++ {
        z := <- y_ch
        fmt.Println("       main y:", i, z)
    }
}

正确产出（变式1）：

main x: 0 [0 0 0 0]
main x: 1 [1 1 1 1]
main x: 2 [2 2 2 2]
main x: 3 [3 3 3 3]
   worker 3 x: [3 3 3 3]
   worker 3 y: [3 3 3 3]
   worker 2 x: [2 2 2 2]
   worker 2 y: [2 2 2 2]
   worker 1 x: [0 0 0 0]
   worker 1 y: [0 0 0 0]
   worker 0 x: [1 1 1 1]
   worker 0 y: [1 1 1 1]
       main y: 0 [3 3 3 3]
       main y: 1 [2 2 2 2]
       main y: 2 [0 0 0 0]
       main y: 3 [1 1 1 1]

输出错误（变量2）：

main x: 0 [0 0 0 0]
main x: 1 [1 1 1 1]
main x: 2 [2 2 2 2]
main x: 3 [3 3 3 3]
   worker 3 x: [3 3 3 3]
   worker 3 y: [3 3 3 3]
       main y: 0 [3 3 3 3]
   worker 0 x: [2 2 2 2]
   worker 0 y: [3 3 3 3]
       main y: 1 [3 3 3 3]
   worker 1 x: [1 1 1 1]
   worker 1 y: [3 3 3 3]
       main y: 2 [3 3 3 3]
   worker 2 x: [3 3 3 3]
   worker 2 y: [3 3 3 3]
       main y: 3 [3 3 3 3]

Answer 1

非常感谢你的解释，现在我知道问题出在哪里了。我添加了一些调试代码来输出指针地址，结果是（稍微重新格式化输出）：

变式1：

main 0 x=[0 0 0 0] &x=0x1830e180 &x[0]=0x1830e1e0
main 1 x=[1 1 1 1] &x=0x1830e180 &x[0]=0x1830e230
main 2 x=[2 2 2 2] &x=0x1830e180 &x[0]=0x1830e270
main 3 x=[3 3 3 3] &x=0x1830e180 &x[0]=0x1830e2a0
    worker 3 x=[3 3 3 3] &x=0x1830e1d0 &x[0]=0x1830e2a0
    worker 3 y=[3 3 3 3] &y=0x1830e2e0 &y[0]=0x1830e2a0
        main 0 y=[3 3 3 3] &y=0x1830e2d0 &y[0]=0x1830e2a0
    worker 0 x=[0 0 0 0] &x=0x1830e1a0 &x[0]=0x1830e1e0
    worker 0 y=[0 0 0 0] &y=0x1830e370 &y[0]=0x1830e1e0
        main 1 y=[0 0 0 0] &y=0x1830e360 &y[0]=0x1830e1e0
    worker 1 x=[1 1 1 1] &x=0x1830e1b0 &x[0]=0x1830e230
    worker 1 y=[1 1 1 1] &y=0x1830e400 &y[0]=0x1830e230
        main 2 y=[1 1 1 1] &y=0x1830e3f0 &y[0]=0x1830e230
    worker 2 x=[2 2 2 2] &x=0x1830e1c0 &x[0]=0x1830e270
    worker 2 y=[2 2 2 2] &y=0x1830e480 &y[0]=0x1830e270
        main 3 y=[2 2 2 2] &y=0x1830e470 &y[0]=0x1830e270

变式2：

main 0 x=[0 0 0 0] &x=0x1830e180 &x[0]=0x1830e190
main 1 x=[1 1 1 1] &x=0x1830e180 &x[0]=0x1830e190
main 2 x=[2 2 2 2] &x=0x1830e180 &x[0]=0x1830e190
main 3 x=[3 3 3 3] &x=0x1830e180 &x[0]=0x1830e190
    worker 3 x=[3 3 3 3] &x=0x1830e1d0 &x[0]=0x1830e190
    worker 3 y=[3 3 3 3] &y=0x1830e2a0 &y[0]=0x1830e190
        main 0 y=[3 3 3 3] &y=0x1830e290 &y[0]=0x1830e190
    worker 0 x=[3 3 3 3] &x=0x1830e1a0 &x[0]=0x1830e190
    worker 0 y=[3 3 3 3] &y=0x1830e330 &y[0]=0x1830e190
        main 1 y=[3 3 3 3] &y=0x1830e320 &y[0]=0x1830e190
    worker 1 x=[3 3 3 3] &x=0x1830e1b0 &x[0]=0x1830e190
    worker 1 y=[3 3 3 3] &y=0x1830e3c0 &y[0]=0x1830e190
        main 2 y=[3 3 3 3] &y=0x1830e3b0 &y[0]=0x1830e190
    worker 2 x=[3 3 3 3] &x=0x1830e1c0 &x[0]=0x1830e190
    worker 2 y=[3 3 3 3] &y=0x1830e440 &y[0]=0x1830e190
        main 3 y=[3 3 3 3] &y=0x1830e430 &y[0]=0x1830e190

Answer 2

不同的是，在变体1中，您每次都发送不同的切片，而在变体2中，您每次都发送相同的切片（在循环的上面创建的切片）。在不创建新切片的情况下，您只是将同一切片的元素设置为不同的值，因此goroutine在查看切片时会看到切片中的任何值。在变体2中，main将始终看到[3 3 3]，因为这是经过4次循环后的最终值。切片对象的值包含对基础元素的引用，而不是元素本身。这里对切片有一个很好的解释。