Question

我正在研究一个霍夫曼代码生成器。下面是我组成树的功能。树基于对象指针的向量。我检查过，似乎工作正常。我现在想把指针传递到位置pointerVect [0]，这应该是树的根到我下面的哈夫曼编码递归函数，但由于某种原因它不能正常工作，就像我试图打印的内容一样存储代码的地图没有打印出来。

class asciiChar  //Individual character module >>> Base Class
{

public:

    void setCharValue (char letter)
    {
        charValue = letter;
    }

    char getCharValue ()
    {
        return charValue;
    }

    void incrementCharCount ()
    {
        charCount++;
    }

    int getCharCount()
    {
        return charCount;
    }

    virtual asciiChar * getLeft()
    {
        return left;
    }

    virtual asciiChar * getRight()
    {
        return right;
    }


    asciiChar(char c, int f)  //Constructor
    {
        charValue = c;
        charCount = f;
    }


    asciiChar & operator= (const asciiChar & other)  //Overloaded assignment operator
    {
        charValue = other.charValue;
        charCount = other.charCount;

        return *this;
    }


    char charValue;
    int charCount = 0;
    asciiChar * left = NULL;
    asciiChar * right = NULL;
};


class parentNode : public asciiChar  //Connector node
{

public:

    parentNode(asciiChar c0, asciiChar c1) : asciiChar(NULL, c0.getCharCount() + c1.getCharCount())
    {
        left = &c0;
        right = &c1;

    }

    ~parentNode()
    {
        if (left) delete left;
        if (right) delete right;
    }

};


asciiChar* createTree (vector<asciiChar> sortedVector)
{
    vector<asciiChar*> pointerVect;
    pointerVect.reserve(sortedVector.size());

    for(int i=0; i < sortedVector.size(); i++)
    {
        pointerVect.push_back(new asciiChar(sortedVector[i].getCharValue(), sortedVector[i].getCharCount()));

    }

    while (pointerVect.size() > 1)
    {
        asciiChar * newL = pointerVect.back();
        pointerVect.pop_back();

        asciiChar * newR = pointerVect.back();
        pointerVect.pop_back();

        asciiChar * parent = new parentNode(* newL, * newR);
        pointerVect.push_back(parent);

        vectSort2 (pointerVect);

    }

    return pointerVect[0]; //Returns pointer at very top (The root of the tree)
}

Answer 1

我怀疑你的第一个功能'createTree'

正如我的初始评论所示，出于各种原因，您应该考虑使用优先级队列。以下是我注意到的问题的快速列表

您正在排序指针向量。因此，指针将根据它们的地址值进行排序，而不是它们指向的对象。但是，您可以提供比较器。如果是这种情况，请忽略此子弹。
每次循环迭代时求向量为O（nLog（n）），其中插入优先级队列并维护排序顺序为O（Log（n））
由于您正在对指针进行排序，因此向量的索引0不能保证是树的根。

考虑使用优先级队列：在头文件

中

 #include <queue>

// Comparator for priority queue. Use this so it compared what the pointers point too  and not the pointers themselves. This way the frequencies are used for the
// comparisons. This forces the priority queue to order from lowest freq
// to the highest frequency
struct CompareHuffChars : public binary_function<asciiChar*, asciiChar*, bool>
{
    bool operator()(const asciiChar* left, const asciiChar* right) const
    {
        // Be sure to add functionality to get frequency for each asciiChar object
        return left->getFrequency() > right->getFrequency();
    }
}; // end struct

priority_queue<asciiChar*,vector<asciiChar*>,CompareHuffChars > * bytePriorityQueue;
asciiChar * huffmanTree; // Pointer to assign to root node of tree when found

在实施文件....

while (!(this->bytePriorityQueue->empty())) {
    asciiChar * qtop = this->bytePriorityQueue->top();
    this->bytePriorityQueue->pop();
if (this->bytePriorityQueue->empty()) {
        // Found the root asciiChar node
        this->huffmanTree = qtop; // huffManTree = asciiChar *
    } else {
        // There are more asciiChar nodes so we need to grab the 2nd from top
        // and combine their frequencies into a new asciiChar node and insert it
        // back into the priority queue

        asciiChar * newNode;
        asciiCharChar * qtopSecond = this->bytePriorityQueue->top();

        // Remove it from the queue
        this->bytePriorityQueue->pop();

        // Now create a new asciiChar node with the added frequences
        // qtopSecond should always be > or = qtop
        // which will adhere to the binary tree structure

        // This assumes asciiChar adds the frequencies of qtop and qtopSecond in constructor
        newNode = new asciiChar(qtop,qtopSecond);

        // Push the new node into the p queue
        // Stays sorted with Log(n) insertion
        this->bytePriorityQueue->push(newNode);

        // Now repeat this until the tree is formed (1 node left in queue)

    } // end if

} // end while

//The p queue should now be completely empty (len =0)

}

现在我的版本需要对asciiChar进行一些重构。但是，此方法应该比发布的方法更好地工作并解决您的错误。

修改

好吧，我想'我发现了你的错误。在asciiChar的头文件中，getLeft和getRight函数是非虚拟。这意味着当你有一个asciiChar *类型的基指针指向一个parentNode（子类）类型的对象时，它将调用父的（asciiChar）getLeft和getRight函数，该函数将始终返回NULL。您在子类（parentNode）中声明了左和右，您不需要这样做，因为这些成员变量在您的父类中是公共的。使getLeft和getRight函数成为虚函数，并删除parentNode类中左右声明及其各自的getter函数。

// In aschiiChar virtual asciiChar * getLeft() { return left; } virtual asciiChar * getRight() { return right; }

备注：在删除之前，您应该检查析构函数是否为NULL。

if (left) delete left; if (right) delete right;

最终修改

感谢您发布更多信息。好吧，你的问题归结为以下几点：

// This is your parentNode constructor parentNode(asciiChar c0, asciiChar c1) : asciiChar(NULL, c0.getCharCount() + c1.getCharCount()) { left = &c0; right = &c1; } // This is what the parentNode constructor should look like parentNode(asciiChar * c0, asciiChar * c1) : asciiChar(NULL, c0->getCharCount() + c1->getCharCount()) { left = c0; right = c1; }

最后......

asciiChar* createTree (vector<asciiChar> sortedVector) { vector<asciiChar*> pointerVect; pointerVect.reserve(sortedVector.size()); for(int i=0; i < sortedVector.size(); i++) { pointerVect.push_back(new asciiChar(sortedVector[i].getCharValue(), sortedVector[i].getCharCount())); } while (pointerVect.size() > 1) { asciiChar * newL = pointerVect.back(); pointerVect.pop_back(); asciiChar * newR = pointerVect.back(); pointerVect.pop_back(); // CHANGE HERE // Don't dereference the pointers. If you dereference them you are passing by value // and creating copies in the constructor which are destroyed upon exit of the constructor asciiChar * parent = new parentNode( newL, newR); pointerVect.push_back(parent); vectSort2 (pointerVect); } return pointerVect[0]; //Returns pointer at very top (The root of the tree) }

你的问题归结为你传递了值并将本地副本的地址分配给parentNode的成员变量指针。然后，parentNode中的这些指针指向不存在的不存在的内存或内存。

希望这有助于......

霍夫曼编码器 - 递归，编码功能失败

1 个答案: