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我想从对象外部操作链接列表，但它不像我想的那样工作。

情况是这样的：我有一个对象，它有一个基类的指针，指向第一个条目，以标记链表的开头。

class theObject {
public:
    theObject() : mFirstEntry(0), mLastEntry(0) {}
    ~theObject() {}

    template<class T>
    void addEntry(const std::string &blah, const std::string &blub, const T &hui)
    {
        child<T> *newEntry = new child<T>(blah, blub, hui);

        if (mFirstEntry) {
            mLastEntry->setNext(newEntry);
            mLastEntry = newEntry;
        }
        else {
            mFirstEntry = newEntry;
            mLastEntry = newEntry;
        }
    }

    base * getFirstEntry() const
    {
        return mFirstEntry;
    }

    void printEntrys() const
    {
        base *data = mFirstEntry;
        while(data) {
            std::cout << data->getBlah() << data->getBlub() << std::endl;
            data = data->getNext();
        }
    }

private:
    base *mFirstEntry;
    base *mLastEntry;
};

class base {
public:
    base() : mBlah(""), mBlub(""), mNext(0) {}

    base(const std::string &blah, const std::string &blub) : mBlah(blah), mBlub(blub), mNext(0) {}

    virtual ~base()
    {
        if (mNext) {
            delete mNext;
        }
    }

    void setNext(base *next)
    {
        mNext = next;
    }

    base * getNext() const
    {
        return mNext;
    }

    std::string getBlah() const
    {
        return mBlah;
    }

    std::string getBlub() const
    {
        return mBlub;
    }

protected:
    std::string mBlah;
    std::string mBlub;
    base *mNext;
};

链表的每个条目都是子类型，它是一个模板并继承了基类。

template<class T>
class child : public base {
public:
    inline child(const std::string &blah, const std::string &blub, const T &hui, base *next = 0) : mHui(hui), base(blah, blub)
    {
        if(next) {
            mNext = next;
        }
    }

    inline child(const child &r)
    {
        *this = r;
    }

    inline const child & operator = (const child &r)
    {
        if (this == &r) return *this;

        mBlah = r.mBlah;
        mBlub = r.mBlub;
        mNext = r.mNext;
        mHui = r.mHui;

        return *this;
    }

    inline const T getData() const
    {
        return mHui;
    }

protected:
    T mHui;
};

现在我用几个条目填充对象

int main(int argc, char* argv[])
{
    theObject data;
    int a(0), b(1), c(2), d(3);
    const std::string blah("blah"), blub("blub");
    data.addEntry(blah, blub, a);
    data.addEntry(blah, blub, b);
    data.addEntry(blah, blub, c);
    data.addEntry(blah, blub, d);

    std::cout << "Original entries" << std::endl;
    data.printEntrys();

然后我想maipulate链表

    base *stuff = data.getFirstEntry();
    std::cout << "Changed in stuff list" << std::endl;
    while(stuff) {
        stuff = new child<double>("noBlah", "noBlub", 3.14, stuff->getNext());
        std::cout << stuff->getBlah() << stuff->getBlub() << std::endl;
        stuff = stuff->getNext();
    }

并希望操纵原件……但它接缝我只操纵了一个副本

    std::cout << "linked list in data object should now be stuff list" << std::endl;
    data.printEntrys();

    return 0;
}

有没有人知道为什么它不工作？getFirstEntry() 返回一个指针，所以我想我会操纵它指向的对象。

最好的问候，本

4

3 回答 3

0

您通过 获得指向第一个元素的指针base *stuff = data.getFirstEntry();，然后将其指向新对象而不是stuff = new child<double>("noBlah", "noBlub", 3.14, stuff->getNext());. 换句话说，您更改的是指针，而不是指向的值。

我建议您查看有关指针的教程。曾经有人推荐过这个视频：Binky Pointer Fun

于 2012-11-13T17:55:10.500 回答
0

好的，感谢 ntor，我能够解决这个问题。

这是代码示例（开头的代码仍然有效）

template<class T>
child<T> * getnewEntry(const std::string &a, const std::string &b, const T &c, base *next)
{
    child<T> *nc= new child<T>(a, b, c, next);
    return nc;
}

int main(int argc, char* argv[])
{
    theObject data;
    int a(0), b(1), c(2), d(3);
    const std::string blah("blah"), blub("blub");
    data.addEntry(blah, blub, a);
    data.addEntry(blah, blub, b);
    data.addEntry(blah, blub, c);
    data.addEntry(blah, blub, d);

    base *stuff = data.getFirstEntry();
    stuff = getnewEntry("noBlah", "noBlub", 3.14, stuff->getNext());
    data.mFirstEntry->setNext(0);
    delete data.mFirstEntry;
    data.mFirstEntry = stuff;

    while (stuff->getNext()) {
        base *tmpstuff = getnewEntry("noBlah", "noBlub", 3.14, stuff->getNext()->getNext());
        stuff->getNext()->setNext(0);
        delete stuff->getNext();
        stuff->setNext(tmpstuff);
        stuff = tmpstuff;
    }

    data.mLastEntry = stuff;

    return 0;
}

这现在正在更改链表位置的对象并更新链表

于 2012-11-14T10:59:22.617 回答
0

您返回一个指针并更改本地指针，而不更改指针指向的值！你应该做：

*stuff = child<double>("noBlah", "noBlub", 3.14, stuff->getNext());

通过这种方式，您实际上会更改指针指向的对象。

LG ntor

于 2012-11-13T17:59:35.807 回答