英文:
Golang Goroutine leak
问题
这段代码是关于Go语言并发的一个解决方案。在Same()函数中，有一个注释说它会泄漏goroutine，这是什么意思呢？它还提到了一个修复这个问题的第二个文件。
在第一个文件中，Same()函数使用了两个goroutine来遍历两个树，并将它们的值发送到两个通道中。然后，通过从通道中接收值并比较它们，判断两个树是否包含相同的值。但是，如果两个树的结构不同，其中一个树的遍历会先结束，而另一个树的goroutine仍然在运行。这就是所谓的goroutine泄漏，因为没有正确地关闭goroutine。
为了解决这个问题，第二个文件中的代码对Walk()函数进行了修改。在walkImpl()函数中，使用了一个额外的quit通道来通知goroutine停止运行。当遍历到树的某个节点时，通过select语句同时监听ch和quit通道，如果quit通道有值可接收，就立即返回，停止继续遍历。这样，即使两个树的结构不同，也能正确地停止遍历并关闭所有的goroutine。
希望这能帮到你！如果还有其他问题，请随时提问。
英文:
// Copyright 2012 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.
// +build ignore
package main
import (
"fmt"
"code.google.com/p/go-tour/tree"
)
func walkImpl(t *tree.Tree, ch chan int) {
if t == nil {
return
}
walkImpl(t.Left, ch)
ch <- t.Value
walkImpl(t.Right, ch)
}
// Walk walks the tree t sending all values
// from the tree to the channel ch.
func Walk(t *tree.Tree, ch chan int) {
walkImpl(t, ch)
// Need to close the channel here
close(ch)
}
// Same determines whether the trees
// t1 and t2 contain the same values.
// NOTE: The implementation leaks goroutines when trees are different.
// See binarytrees_quit.go for a better solution.
func Same(t1, t2 *tree.Tree) bool {
w1, w2 := make(chan int), make(chan int)
go Walk(t1, w1)
go Walk(t2, w2)
for {
v1, ok1 := <-w1
v2, ok2 := <-w2
if !ok1 || !ok2 {
return ok1 == ok2
}
if v1 != v2 {
return false
}
}
}
func main() {
fmt.Print("tree.New(1) == tree.New(1): ")
if Same(tree.New(1), tree.New(1)) {
fmt.Println("PASSED")
} else {
fmt.Println("FAILED")
}
fmt.Print("tree.New(1) != tree.New(2): ")
if !Same(tree.New(1), tree.New(2)) {
fmt.Println("PASSED")
} else {
fmt.Println("FAILED")
}
}
In this code, a solution for http://tour.golang.org/concurrency/8
Why is there a comment on Same() func Same(t1, t2 *tree.Tree) bool saying that it leaks goroutines? How so? It also mentions a second file that fixes this:
// Copyright 2015 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.
// +build ignore
package main
import (
"fmt"
"code.google.com/p/go-tour/tree"
)
func walkImpl(t *tree.Tree, ch, quit chan int) {
if t == nil {
return
}
walkImpl(t.Left, ch, quit)
select {
case ch <- t.Value:
// Value successfully sent.
case <-quit:
return
}
walkImpl(t.Right, ch, quit)
}
// Walk walks the tree t sending all values
// from the tree to the channel ch.
func Walk(t *tree.Tree, ch, quit chan int) {
walkImpl(t, ch, quit)
close(ch)
}
// Same determines whether the trees
// t1 and t2 contain the same values.
func Same(t1, t2 *tree.Tree) bool {
w1, w2 := make(chan int), make(chan int)
quit := make(chan int)
defer close(quit)
go Walk(t1, w1, quit)
go Walk(t2, w2, quit)
for {
v1, ok1 := <-w1
v2, ok2 := <-w2
if !ok1 || !ok2 {
return ok1 == ok2
}
if v1 != v2 {
return false
}
}
}
func main() {
fmt.Print("tree.New(1) == tree.New(1): ")
if Same(tree.New(1), tree.New(1)) {
fmt.Println("PASSED")
} else {
fmt.Println("FAILED")
}
fmt.Print("tree.New(1) != tree.New(2): ")
if !Same(tree.New(1), tree.New(2)) {
fmt.Println("PASSED")
} else {
fmt.Println("FAILED")
}
}
How does it accomplish that? Where was this leak? (to test the code you will have to run it on http://tour.golang.org/concurrency/8). Very confused and would appreciate some help, thanks!
答案1
得分: 2
程序在检测到差异时停止接收通道上的数据。
walk协程会一直运行，直到它们被阻塞在发送数据到通道上。它们永远不会退出。这就是泄漏的原因。
英文:
The program stops receiving on the channels when a difference is detected.
The walk goroutines run until they block sending to the channels. They never exit. This is the leak.
答案2
得分: 0
第二种解决方案通过使用退出通道来处理泄漏。当退出通道关闭时（在Same（）函数中），选择语句的第2个case成功（在walkImpl函数中的case <-quit），并且函数返回。因此，在程序退出后，walkImpl函数中没有阻塞。
英文:
The second solution handles the leak by using the quit channel. When the quit channel is closed (in Same() function), the case 2 of select statement succeeds(case <-quit in walkImpl function) and the function returns. Hence there is no block in the walkImpl function after the program exits.
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