用另一個對象作爲參數在構造函數中創建一個對象

Question

因此，我正在製作一個程序，它將查找所有可能的單詞（由字典定義）的一個boggle遊戲（谷歌它，如果你不知道它是什麼是）。無論如何，我有3類/對象：字典，遊戲板和bogglegame。 bogglegame應該梳理字典和遊戲板，並找到所有合法的單詞。我對bogglegame構造看起來像

BoggleGame::BoggleGame(Dictionary dictionaryIN, GameBoard gameboardIN)

和詞典和遊戲鍵盤的contsrtuctors樣子

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

，當我嘗試編譯我得到，說：「錯誤的錯誤：沒有匹配函數調用'Dictionary :: Dictionary（）「

我想能夠通過在詞典ry和gameboard對象從主存入到構造函數中，並將它們存儲爲BoggleGame對象的私有成員，使BoggleGame對象成爲2個對象的對象。

編輯：郵政代碼

構造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

錯誤在於BoggleGame的構造函數。您沒有Dictionary的默認構造函數。你可以代替你的構造使用初始化列表：

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

或者，Dictionary提供一個默認的構造函數：

Dictionary() = default; 
Dictionary() { } 

---

出現錯誤的原因是因爲dictionary是缺省構造的。當您將新值分配給dictionary時，您正在使用複製分配操作員。通過使用構造函數初始化列表，您可以直接初始化dictionary。爲了演示：

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

Default. 
Copy assignment. 

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

Copy constructor. 

大的差異，你可以看到。

Answer 2

也許這是您在類BoggleGame中聲明Dictionary對象的方式。

比方說，你做某事像這樣：

類BoggleGame {

私人： 字典迪科;

}

在這種情況下，會出現一個編譯錯誤，如果構造字典::字典（）不存在。

所以我覺得，你的問題的解決方案應該是兩種：

• 聲明無參數的構造函數中Dictionary類
• 申報迪科爲指針（也就是平均字典*迪科）

但是，如果您發佈代碼會更好。