Golang 协程泄漏
Golang Goroutine leak
// 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")
}
}
在此代码中,http://tour.golang.org/concurrency/8
的解决方案
为什么 Same() func Same(t1, t2 *tree.Tree) bool 上有评论说它泄漏了 goroutines?怎么会这样?它还提到了修复此问题的第二个文件:
// 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")
}
}
它是如何实现的?这个泄漏在哪里? (要测试代码,您必须 运行 在 http://tour.golang.org/concurrency/8 上)。非常困惑,希望得到一些帮助,谢谢!
程序在检测到差异时停止在频道上接收。
walk goroutines 运行 直到它们阻止发送到通道。他们从不退出。这就是漏洞。
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),select 语句的 case 2 成功(在 walkImpl 函数中的 case <-quit)和函数 returns。因此程序退出后walkImpl函数中没有block。
// 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")
}
}
在此代码中,http://tour.golang.org/concurrency/8
的解决方案为什么 Same() func Same(t1, t2 *tree.Tree) bool 上有评论说它泄漏了 goroutines?怎么会这样?它还提到了修复此问题的第二个文件:
// 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")
}
}
它是如何实现的?这个泄漏在哪里? (要测试代码,您必须 运行 在 http://tour.golang.org/concurrency/8 上)。非常困惑,希望得到一些帮助,谢谢!
程序在检测到差异时停止在频道上接收。
walk goroutines 运行 直到它们阻止发送到通道。他们从不退出。这就是漏洞。
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),select 语句的 case 2 成功(在 walkImpl 函数中的 case <-quit)和函数 returns。因此程序退出后walkImpl函数中没有block。