用另一个对象作为参数在构造函数中创建一个对象

Question

因此，我正在制作一个程序，它将查找所有可能的单词（由字典定义）的一个boggle游戏（谷歌它，如果你不知道它是什么是）。无论如何，我有3类/对象：字典，游戏板和bogglegame。 bogglegame应该梳理字典和游戏板，并找到所有合法的单词。我对bogglegame构造看起来像

BoggleGame::BoggleGame(Dictionary dictionaryIN, GameBoard gameboardIN)

和词典和游戏键盘的contsrtuctors样子

Dictionary::Dictionary(set<string> wordsInDictionaryIN, unsigned maxLengthIN) GameBoard::GameBoard(vector<vector<string> > gamestateIN, unsigned boardSizeIN)

，当我尝试编译我得到，说：“错误的错误：没有匹配函数调用'Dictionary :: Dictionary（）“

我想能够通过在词典ry和gameboard对象从主存入到构造函数中，并将它们存储为BoggleGame对象的私有成员，使BoggleGame对象成为2个对象的对象。

编辑：邮政代码

构造BoggleGame

#include "BoggleGame.h" 

BoggleGame::BoggleGame(Dictionary dictionaryIN, GameBoard gameboardIN) 
{ 
    dictionary = dictionaryIN; 
    gameboard = gameboardIN; 
} 

---

#pragma once 

#include "Dictionary.h" 
#include "GameBoard.h" 

class BoggleGame 
{ 
public: 
    BoggleGame(Dictionary dictioanryIN, GameBoard gameboardIN); 
    void foundWord(string wordIN); 
    string findTheWords(string w, unsigned row, unsigned column); 
    set<string> getTheFoundWords() {return foundWords;} 
    bool isInDictionary(string word); 
    bool isOffBoard(unsigned row, unsigned column); 
    bool usedTile(unsigned row, unsigned column); 
    vector<vector<string> > getTheGameBoard(){gameboard.getGameBoard();} 

private: 
    Dictionary dictionary; 
    GameBoard gameboard; 
    set<string> foundWords; 
}; 

---

#pragma once 
#include <set> 
#include <vector> 
#include <string> 
#include <iterator> 

using std::set; 
using std::string; 
using std::vector; 

class Dictionary 
{ 
public: 
    Dictionary(set<string> wordsInDictionaryIN, unsigned maxLengthIN); 
    bool isInDictionary(string wordIN); 
    void foundWord (string wordIN); 
    string findTheWords(string w, unsigned row, unsigned column); 
    unsigned getMaxLength() {return maxLength;} 

private: 
    set<string> wordsInDictionary; 
    unsigned maxLength; 
    set<string> foundWords; 

}; 

---

#pragma once 
#include <vector> 
#include <string> 

using std::string; 
using std::vector; 

class GameBoard 
{ 
public: 
    GameBoard(vector<vector<string> > gamestateIN, unsigned boardSizeIN); 
    bool outOfBoard(unsigned row, unsigned column); 
    bool getTileUseState(unsigned row, unsigned column){return usedTiles.at(row).at(column);} 
    void setTileUsed(unsigned row, unsigned column); 
    vector<vector<string> > getGameBoard(){return gamestate;} 
    unsigned getSize(){return boardSize;} 
    vector<vector<bool> > getUsedTiles() {return usedTiles;} 
    string readTile(unsigned row, unsigned column) {return gamestate.at(row).at(column);} 
    void previousState(vector<vector<bool> > previous) {usedTiles = previous;} 

private: 
    vector<vector<string> > gamestate; 
    vector<vector<bool> > usedTiles; 
    unsigned boardSize; 
}; 

Answer 1

错误在于BoggleGame的构造函数。您没有Dictionary的默认构造函数。你可以代替你的构造使用初始化列表：

BoggleGame::BoggleGame(Dictionary _dictionary) 
    : dictionary{_dictionary} 

或者，Dictionary提供一个默认的构造函数：

Dictionary() = default; 
Dictionary() { } 

---

出现错误的原因是因为dictionary是缺省构造的。当您将新值分配给dictionary时，您正在使用复制分配操作员。通过使用构造函数初始化列表，您可以直接初始化dictionary。为了演示：

BoggleGame::BoggleGame(Dictionary _dictionary) 
{ 
    dictionary = _dictionary; 
} 

Default. 
Copy assignment. 

BoggleGame::BoggleGame(Dictionary _dictionary) 
    : dictionary{_dictionary} 
    { 
    } 

Copy constructor. 

大的差异，你可以看到。

Answer 2

也许这是您在类BoggleGame中声明Dictionary对象的方式。

比方说，你做某事像这样：

类BoggleGame {

私人： 字典迪科;

}

在这种情况下，会出现一个编译错误，如果构造字典::字典（）不存在。

所以我觉得，你的问题的解决方案应该是两种：

• 声明无参数的构造函数中Dictionary类
• 申报迪科为指针（也就是平均字典*迪科）

但是，如果您发布代码会更好。