使用 select 的 Golang 频道不会停止

Golang channels using select doesn't stop

这里是 Go-lang 新手。我正在尝试 Go 的 Tour of Go,遇到了一个关于通道的练习 (https://tour.golang.org/concurrency/7)。 这个想法是走两棵树,然后评估树是否等价。

我想使用 select 等待来自两个渠道的结果来解决这个练习。当两者都完成时,我评估生成的切片。不幸的是,该方法进入无限循环。我添加了一些输出以查看发生了什么,并注意到只有一个通道被关闭,然后再次打开。

我明明做错了什么,但我看不出是什么。 我的问题是我做错了什么?关于关闭使下面的代码进入无限循环的通道,我做了什么假设?

package main

import (
    "golang.org/x/tour/tree"
    "fmt"
)

// Walk walks the tree t sending all values
// from the tree to the channel ch.
func Walk(t *tree.Tree, ch chan int) {
    _walk(t, ch)
    close(ch)
}

func _walk(t *tree.Tree, ch chan int) {
    if (t.Left != nil) {
        _walk(t.Left, ch)
    }
    ch <- t.Value
    if (t.Right != nil) {
        _walk(t.Right, ch)
    }
}

// Same determines whether the trees
// t1 and t2 contain the same values.
func Same(t1, t2 *tree.Tree) bool {
    ch1 := make(chan int)
    ch2 := make(chan int)
    go Walk(t1, ch1)
    go Walk(t2, ch2)

    var out1 []int
    var out2 []int

    var tree1open, tree2open bool
    var tree1val, tree2val int
    for {
        select {
        case tree1val, tree1open = <- ch1:
            out1 = append(out1, tree1val)
        case tree2val, tree2open = <- ch2:
            out2 = append(out2, tree2val)
        default:
            if (!tree1open && !tree2open) {
                break
            } else {
                fmt.Println("Channel open?", tree1open, tree2open)
            }
        }
    }

    if (len(out1) != len(out2)) {
        return false
    }

    for i := 0 ; i < len(out1) ; i++ {
        if (out1[i] != out2[i]) {
            return false
        }
    }

    return true
}

func main() {
    ch := make(chan int)
    go Walk(tree.New(1), ch)

    for i := range ch {
        fmt.Println(i)
    }

    fmt.Println(Same(tree.New(1), tree.New(1)))
    fmt.Println(Same(tree.New(1), tree.New(2)))
}

A "break" 语句终止执行最里面的 "for"、"switch" 或 "select" 语句。
参见:http://golang.org/ref/spec#Break_statements
您示例中的 break 语句终止 select 语句,即 "innermost" 语句。
所以添加label: ForLoop before for循环并添加break ForLoop

ForLoop:
    for {
        select {
        case tree1val, tree1open = <-ch1:
            if tree1open {
                out1 = append(out1, tree1val)
            } else if !tree2open {
                break ForLoop
            }
        case tree2val, tree2open = <-ch2:
            if tree2open {
                out2 = append(out2, tree2val)
            } else if !tree1open {
                break ForLoop
            }
        }
    }

如果您想自己解决该问题,请不要阅读其余部分,完成后再回来: 解决方案 1(与您的类似):

package main

import "fmt"
import "golang.org/x/tour/tree"

// Walk walks the tree t sending all values
// from the tree to the channel ch.
func Walk(t *tree.Tree, ch chan int) {
    _walk(t, ch)
    close(ch)
}

func _walk(t *tree.Tree, ch chan int) {
    if t.Left != nil {
        _walk(t.Left, ch)
    }
    ch <- t.Value
    if t.Right != nil {
        _walk(t.Right, ch)
    }
}

// Same determines whether the trees
// t1 and t2 contain the same values.
func Same(t1, t2 *tree.Tree) bool {
    ch1, ch2 := make(chan int), make(chan int)
    go Walk(t1, ch1)
    go Walk(t2, ch2)

    tree1open, tree2open := false, false
    tree1val, tree2val := 0, 0
    out1, out2 := make([]int, 0, 10), make([]int, 0, 10)
ForLoop:
    for {
        select {
        case tree1val, tree1open = <-ch1:
            if tree1open {
                out1 = append(out1, tree1val)
            } else if !tree2open {
                break ForLoop
            }
        case tree2val, tree2open = <-ch2:
            if tree2open {
                out2 = append(out2, tree2val)
            } else if !tree1open {
                break ForLoop
            }
        }
    }
    if len(out1) != len(out2) {
        return false
    }
    for i, v := range out1 {
        if v != out2[i] {
            return false
        }
    }
    return true
}

func main() {
    ch := make(chan int)
    go Walk(tree.New(1), ch)
    for i := range ch {
        fmt.Println(i)
    }
    fmt.Println(Same(tree.New(1), tree.New(1)))
    fmt.Println(Same(tree.New(1), tree.New(2)))
}

输出:

1
2
3
4
5
6
7
8
9
10
true
false    

另一种方式:

package main

import "fmt"
import "golang.org/x/tour/tree"

// Walk walks the tree t sending all values
// from the tree to the channel ch.
func Walk(t *tree.Tree, ch chan int) {
    _walk(t, ch)
    close(ch)
}

func _walk(t *tree.Tree, ch chan int) {
    if t != nil {
        _walk(t.Left, ch)
        ch <- t.Value
        _walk(t.Right, ch)
    }
}

// Same determines whether the trees
// t1 and t2 contain the same values.
func Same(t1, t2 *tree.Tree) bool {
    ch1, ch2 := make(chan int), make(chan int)
    go Walk(t1, ch1)
    go Walk(t2, ch2)
    for v := range ch1 {
        if v != <-ch2 {
            return false
        }
    }
    return true
}

func main() {
    ch := make(chan int)
    go Walk(tree.New(1), ch)
    for v := range ch {
        fmt.Println(v)
    }
    fmt.Println(Same(tree.New(1), tree.New(1)))
    fmt.Println(Same(tree.New(1), tree.New(2)))
}

输出:

1
2
3
4
5
6
7
8
9
10
true
false    

并查看:
Go Tour Exercise: Equivalent Binary Trees

AMD的建议在之前的回答中是有效的。但是,看看您要解决的问题,它仍然没有解决它。 (如果你 运行 程序,它会在两种情况下输出 true)

这是问题所在:

for {
    select {
    case tree1val, tree1open = <-ch1:
        out1 = append(out1, tree1val)
    case tree2val, tree2open = <-ch2:
        out2 = append(out2, tree2val)
    default:
        //runtime.Gosched()
        if !tree1open && !tree2open {
            break ForLoop
        } else {
            fmt.Println("Channel open?", tree1open, tree2open)
        }
    }
}

在这种情况下,由于 tree1open 和 tree2open 的默认值是 false(根据 golang 规范),它会转到 'default' 的情况,因为 select 是非阻塞的,并且简单从 ForLoop 中断,甚至没有填充 out1 和 out2 切片(可能,因为它们是 goroutine)。因此 out1 和 out2 的长度保持为零,因此在大多数情况下它输出 true。

更正如下:

ForLoop:
for {
    select {
    case tree1val, tree1open = <-ch1:
        if tree1open {
            out1 = append(out1, tree1val)
        }
        if !tree1open && !tree2open {
            break ForLoop
        }
    case tree2val, tree2open = <-ch2:
        if tree2open {
            out2 = append(out2, tree2val)
        }
        if !tree1open && !tree2open {
            break ForLoop
        }
    default:

    }
}

需要注意的关键是我们必须检查两种情况下通道是否已经关闭(相当于说tree1open和tree2open是否都是假的)。在这里,它会正确填充 out1 和 out2 切片,然后进一步比较它们各自的值。

在附加之前添加了对 tree1open(或 tree2open)是否为真的检查,只是为了避免将零值附加到 out1(或 out2)。

谢谢,