douziqian2871
2015-03-22 02:35 浏览 56
已采纳

Golang Goroutine泄漏

// 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!

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2条回答 默认 最新

  • 已采纳
    dtr84664 dtr84664 2015-03-22 02:42

    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.

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  • duandi8852752 duandi8852752 2016-06-24 09:58

    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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