如何在 C++ 中声明指向具有不同数据类型的类模板指针的指针

Question

我有一个用模板设计的程序来实现霍夫曼树。我的 HuffmanNode 类有两种不同的数据类型。这是类的实现。

#pragma once

template <typename T, typename U> class HuffmanNode 
{
private:
    int frequency;
    T value;
    U code;
    HuffmanNode<T, U>* left, *right;
public:
    HuffmanNode<T, U>() :
        value(' '), frequency(0), left(nullptr), right(nullptr) {}
    HuffmanNode<T, U>(T v, U i) :
        value(v), frequency(i), left(nullptr), right(nullptr) {}
    HuffmanNode<T, U>(HuffmanNode<T, U>* left, HuffmanNode<T, U>* right)
    {
        this->left = left;
        this->right = right;
        frequency = left->getFrequency() + right->getFrequency();
        value = NULL;
    }
};

在我的 HuffTree 类中，我有一个指向 HuffmanNode 指针的指针。我尝试以几种不同的方式声明指向 HuffmanNode 指针的指针，但无论哪种方式，我都会收到错误。

我是这样声明的：

HuffmanNode<T>** storage;   // Error: 'HuffmanNode': too few template arguments

然后我尝试了这两种数据类型：

template <class T, class U> 
HuffmanNode<T, U>** storage;   // Error: 'HuffTree<T>::storage': only static data member templates are allowed

如何声明这样的指针？

编辑：我更新了 HuffTree 类以在 HuffmanNode 指针中包含这两种数据类型，但现在我在包含 HuffTree 类的 HuffmanCode 类中遇到错误。这是 HuffTree 类的更新代码：

#pragma once
#include "HuffmanNode.h"
using namespace std;

template <typename T, typename U> class HuffTree 
{
private:
    HuffmanNode<T, U>** storage;
    int treeSize;
    int capacity;

public:
HuffTree<T, U>(int size)
{
    capacity = size;
    treeSize = 0;
    storage = new HuffmanNode<T, U>*[capacity];
    for (int i = 0; i < capacity; i++)
    {
        storage[i] = NULL;
    }
}
// I get an error in the HuffmanCode class when calling this function
void insert(HuffmanNode<T, U>* rhs)
{
    treeSize++;
    storage[treeSize - 1] = rhs;
    percUp(treeSize - 1);
}
};

这是 HuffmanCode 类的代码，现在我修复了指针问题，它给了我一个错误：

#pragma once
#include <string>
#include <map>
#include"HuffmanTree.h"
using std::string;
using std::map;
template <typename T, typename U> class HuffmanCode 
{
private:
    string data;
    string encodedData;
    HuffTree<T, U>* tree;
    map<T, U> frequencyTable;
    map<T, U> huffmanTable;
    void buildTree()
    {
        for (map<T, U>::iterator it = frequencyTable.begin();
            it != frequencyTable.end(); ++it)
        {
            /* 
            this line gives me an error: 'HuffmanNode<char,std::string>::
            HuffmanNode(const HuffmanNode<char,std::string> &)': 
            cannot convert argument 1 from 'const std::string' to 'char'
            */
            tree->insert(new HuffmanNode<char, string>(it->first, it->second));
        }
    }
public:
    HuffmanCode<T, U>(T input)
    {
        data = input;
        encodedData = "";
        // create tree to be large enough for worst case
        tree = new HuffTree<T, U>(data.length());
        buildTable();
        buildTree();
    }
    };

这是我测试这些功能的源文件：

#include <iostream>
#include <string>
#include "HuffmanCode.h"
using namespace std;

int main()
{
    string text;
    cout << "Enter some text:\n";
    getline(cin, text);
    HuffmanCode<string, string>* huffText = new HuffmanCode<string, string>(text);

    huffText->displayTable();
    huffText->displayHuffmanTable();

    string code = huffText->getEncodedString();
    cout << "Encoded string: " << code << endl << endl;
    cout << "Decoded string: " << huffText->decodeString(code) << endl << endl;

    delete huffText;

    return 0;
}

已编辑以包含相关代码

Answer 1

如果您无法定义其他模板参数类型，请在HuffTree 类中自行指定。其中之一应该适合您：

HuffmanNode<T, T> ** storage;

或

HuffmanNode<T, string> ** storage;

（或您认为合适的任何其他数据类型）