go - 如何创建包装容器/list.List 的 golang 结构？

Question

让我的结构包含 list.List 与 *list.List 之间的意外差异令我抓狂。为什么以下不起作用？

type listHolder struct {
    id     int
    mylist list.List
}

func newListHolder(id int, text string) listHolder {
    var newLH listHolder
    newLH.mylist = *list.New()
    newLH.id = id
    newLH.mylist.PushBack(text)
    return newLH
}

func (l *listHolder) pushBack(text string) {
    l.mylist.PushBack(text)
}

func (l *listHolder) printAll() {
    for temp := l.mylist.Front(); temp != nil; temp = temp.Next() {
        fmt.Println(temp.Value)
    }
}


func main() {
    a := newListHolder(1, "first") 
    a.pushBack("second")
    fmt.Printf("listHolder %d length %d Front()= %v, Back()=%v\n",
         a.id, a.mylist.Len(), a.mylist.Front().Value, a.mylist.Back().Value)
    a.printAll()
}

这会输出以下内容，表明长度符合预期，但 Front() 和 Back() 方法不起作用。

listHolder 1 length 2 Front()= `<nil>`, Back()=<nil>
<nil>

如果我将结构定义为

// Same thing with a pointer
type listHolderPtr struct {
    id     int
    mylist *list.List
}

func newListHolderPtr(id int, text string) listHolderPtr {
    var newLH listHolderPtr
    newLH.mylist = list.New()
    newLH.id = id
    newLH.mylist.PushBack(text)
    return newLH
}

按预期工作，但当然，listHolder 结构的任何副本都共享对同一列表的引用，这不是我想要的。我需要能够复制周围的对象并获得内部列表的新副本。这可能吗？

参见 https://play.golang.org/p/KCtTwuvaS1R有关我正在尝试做的事情的简化示例。在实际用例中，我将在一个复杂的嵌套循环中将每个 listHolder 推到后面并从 listHolder 的前面弹出。

Answer 1

我认为 @JimB 的建议是在将值推送到 listHolder.mylist 时制作列表的本地副本可能是解决问题的好方法(假设我理解底层正确发出)。我试图想出一个如下所示的实现:

package main

import (
    "container/list"
    "fmt"
)

type listHolder struct {
    id     int
    mylist list.List
}

func newListHolder(id int) listHolder { // don't push back when constructing a new listHolder
    var newLH listHolder
    newLH.mylist = *list.New()
    newLH.id = id
    return newLH
}

func (l *listHolder) pushBack(text string) {
    // create a temporary list to copy all old and the new value to
    tmpList := list.New()

    // copy all existing values from l.mylist
    for e := l.mylist.Front(); e != nil; e = e.Next() {
        fmt.Printf("pushing back '%v' from old list\n", e.Value)
        tmpList.PushBack(e.Value)
    }

    // push back the new value
    tmpList.PushBack(text)

    // print the new tmpList for debugging purposes
    for ele := tmpList.Front(); ele != nil; ele = ele.Next() {
        fmt.Printf("creating new list element: %v\n", ele.Value)
    }

    // replace l.mylist with tmpList
    l.mylist = *tmpList
    // another version of this solution could be to return a new (i.e. copied) 
    // *listHolder with all the old values and the new 'text' value
}

func (l *listHolder) printAll() {
    for temp := l.mylist.Front(); temp != nil; temp = temp.Next() {
        fmt.Println(temp.Value)
    }
}

func main() {
    a := newListHolder(1)
    a.pushBack("first")  // push a value to a
    a.pushBack("second") // push another value to a
    fmt.Printf("listHolder %d length %d Front()=%v, Back()=%v\n",
        a.id, a.mylist.Len(), a.mylist.Front().Value, a.mylist.Back().Value)
    a.printAll()
}

这段代码输出:

creating new list element: first                         // nothing to copy, only creating a new list element
pushing back 'first' from old list                       // copy element ...
creating new list element: first                         // ... from old list
creating new list element: second                        // and push new element to 'tmpList'
listHolder 1 length 2 Front()=first, Back()=second       // print a summary
first                                                    // of the
second                                                   // new list

如果我有一些模拟数据，我可以做更多的测试/调试。至少这段代码可以在没有 *list.List 的情况下工作。