使用select的Golang频道不会停止

Question

Go-lang newbie here. I am trying out Go's Tour of Go, and came across an exercise about channels (https://tour.golang.org/concurrency/7). The idea is to walk two trees and then evaluate if the trees are equivalent.

I wanted to solve this exercise using a select waiting for results from both channels. When both would finish I evaluate the resulting slice. Unfortunately the method goes on an infinite loop. I added some output to see what was happening and noticed that only one of the channels was being closed, and then opened again.

I am clearly doing something wrong, but I can't see what. My question is what am I doing wrong? What assumption am I making regarding the closing of channels that makes the code below go into an infinite loop?

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)))
}

Answer 1

A "break" statement terminates execution of the innermost "for", "switch" or "select" statement.
see: http://golang.org/ref/spec#Break_statements
the break statement in your example terminates the select statement, the "innermost" statement.
so add label: ForLoop before for loop and add 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
            }
        }
    }

don't read the rest if you want to solve that problem yourself, and come back when you are done: solution 1 (similar to yours):

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)))
}

output:

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

another way:

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)))
}

output:

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

and see:
Go Tour Exercise: Equivalent Binary Trees