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,
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,