井字游戏使用 C++ 和多维数组 [关闭]
Posted
技术标签:
【中文标题】井字游戏使用 C++ 和多维数组 [关闭]【英文标题】：Tic Tac Toe using C++ and multidimensional arrays [closed] 【发布时间】：2016-04-01 20:17:51 【问题描述】：感谢大家的帮助，我是编程新手，不是很熟练，所以感谢您的耐心等待。我让它工作了！代码中有几个问题，从用于检查“游戏结束”的 for 循环和用于设置棋盘的计数器。但最终的代码似乎没有错误。它在控制台中打印出一个带有标记方块的网格，允许两个人一起玩。现在很开心。
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
bool over = false;
void print_board(char board[3][3])
cout<<"\n------------------\n";
cout<<"\n "<<board[0][0]<<" | "<<board[0][1]<< " | "<<board[0][2];
cout<<"\n------------------\n";
cout<<"\n "<<board[1][0]<<" | "<<board[1][1]<< " | "<<board[1][2];
cout<<"\n------------------\n";
cout<<"\n "<<board[2][0]<<" | "<<board[2][1]<< " | "<<board[2][2];
cout<<"\n------------------\n";
bool checkForCorrectMove (char board[3][3], int choice, bool player1Turn)
int xCor;
int yCor;
xCor = ((choice + 2) / 3) - 1;
yCor = (choice - 1) % 3;
if (((board[xCor][yCor]) != 'X') && ((board[xCor][yCor]) != 'O'))
if (player1Turn)
board[xCor][yCor] = 'X';
if (!player1Turn)
board[xCor][yCor] = 'O';
return true;
else
return false;
bool checkRow(char board[3][3])
for (int i = 0; i<3; i++)
if ((board[i][0] == board[i][1]) && (board[i][1] == board[i][2]))
return true;
return false;
bool checkColumn (char board[3][3])
for (int i = 0; i<3; i++)
if ((board[0][i] == board[1][i]) && (board[1][i] == board[2][i]))
return true;
return false;
bool checkForWin (char board[3][3])
bool win = false;
bool line_win = false;
if ((board[0][0] == board[1][1]) && (board[1][1] == board[2][2]))
win = true;
return win;
if ((board[2][0] == board[1][1]) && (board[1][1] == board[0][2]))
win = true;
return win;
if (checkColumn(board) || checkRow (board))
win = true;
return win;
return win;
int main()
cout << "Welcome to Tic Tac Toe!\n";
int choice;
int boardcounter = 1;
bool isOkayToMove = true;
bool player1Turn = true;
char board[3][3];
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++)
board[i][j] = boardcounter + '0';
boardcounter++;
print_board(board);
do
if (player1Turn)
cout <<"\n\nPlayer 1, please enter a number that corresponds to an open space: ";
else
cout <<"\n\nPlayer 2, please enter a number that corresponds to an open space: ";
cin>>choice;
isOkayToMove = checkForCorrectMove(board, choice, player1Turn);
if (isOkayToMove)
over = checkForWin(board);
print_board(board);
player1Turn = !player1Turn;
else
cout <<"\n\nYou have attempted to move into a space that is already occupied, please try again.";
print_board(board);
while (over == false);
cout <<"\n\nCongratulations!";
player1Turn = !player1Turn;
if (player1Turn)
cout<<" Player 1 has won!";
else
cout<<" Player 2 has won!";
我正在尝试用 C++ 创建一个井字游戏，但我很挣扎。我知道以前有人问过这个问题，但是通过查看其他人的示例很难弄清楚我的特定代码（我正在学习 C++）有什么问题。代码如下：
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
bool over = false;
void print_board(char board[3][3])
cout<<"\n------------------\n";
cout<<"\n "<<board[0][0]<<" | "<<board[0][1]<< " | "<<board[0][2];
cout<<"\n------------------\n";
cout<<"\n "<<board[1][0]<<" | "<<board[1][1]<< " | "<<board[1][2];
cout<<"\n------------------\n";
cout<<"\n "<<board[2][0]<<" | "<<board[2][1]<< " | "<<board[2][2];
cout<<"\n------------------\n";
bool checkForCorrectMove (char board[][3], int choice, bool player1Turn)
int xCor;
int yCor;
switch(choice)
case 1: xCor = 0, yCor = 0;
case 2: xCor = 0, yCor = 1;
case 3: xCor = 0, yCor = 2;
case 4: xCor = 1, yCor = 0;
case 5: xCor = 1, yCor = 1;
case 6: xCor = 1, yCor = 2;
case 7: xCor = 2, yCor = 0;
case 8: xCor = 2, yCor = 1;
case 9: xCor = 2, yCor = 2;
if (((board[xCor][yCor]) != 'X') && ((board[xCor][yCor]) != 'O'))
if (player1Turn)
board[xCor][yCor] = 'X';
if (!player1Turn)
board[xCor][yCor] = 'O';
return true;
else
return false;
bool checkForWin (char board[][3])
bool win = false;
if ((board[0][0] == board[1][1]) && (board[1][1] == board[2][2]))
win = true;
else if ((board[2][0] == board[1][1]) && (board[1][1] == board[0][2]))
win = true;
for (int i = 0; i<3; i++)
for (int j = 0; i < 3; j++)
if (board[i][j] != board [i][0]);
win = false;
for (int i = 0; i<3; i++)
for (int j = 0; i < 3; j++)
if (board[i][j] != board [0][j]);
win = false;
return win;
int main()
cout << "Welcome to Tic Tac Toe!\n";
int choice;
int boardcounter = 1;
bool isOkayToMove = true;
bool player1Turn = true;
char board[3][3];
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++)
board[i][j] = (char) boardcounter;
boardcounter++;
print_board(board);
do
if (player1Turn)
cout <<"\n\nPlayer 1, please enter a number that corresponds to an open space: ";
else
cout <<"\n\nPlayer 2, please enter a number that corresponds to an open space: ";
cin>>choice;
switch(choice);
isOkayToMove = checkForCorrectMove(board, choice, player1Turn);
if (isOkayToMove)
print_board(board);
over = checkForWin(board);
player1Turn = !player1Turn;
else
cout <<"\n\nYou have attempted to move into a space that is already occupied, please try again.";
print_board(board);
while (over == false);
cout <<"\n\nCongratulations!";
if (player1Turn)
cout<<" Player 1 has won!";
else
cout<<" Player 2 has won!";
【问题讨论】：
“我在挣扎”是一个不合适的问题描述。我可以告诉你，你的switch 在每个case 上都缺少breaks（它们本身有一些有问题的赋值语句——你从哪里听说C++ 用逗号分隔语句？）但是，除此之外，你要去需要提供一个 MCVE 来解释实际问题是什么......
那有什么问题呢？如果它没有编译，您应该发布错误消息。如果它没有按预期运行，请说明您的预期和实际得到的结果。
@JonathanMa - 你的switch可以压缩成xCor = ((choice + 2) / 3) - 1; yCor = choice % 3;
它可以编译——但现在，问题在于 print_board 函数以及我如何初始化板。它打印两行乱码和第三行空。
您可以使用for 语句来打印板。循环结构对数组（包括二维数组）非常有用。
【参考方案1】：
这是我发现的一个问题：
int boardcounter = 1;
//...
for (int j = 0; j < 3; j++)
board[i][j] = (char) boardcounter;
boardcounter++;
问题在于演员(char) 不会将int 变量转换为文本 表示。强制转换实际上将整数转换为更小的整数变量。
有很多方法可以将数字转换为字符，例如snprintf、tostring 和ostringstream。
由于您的范围有限，即数字 0 - 9，您可能能够逃脱：
board[row][column] = '0' + boardcounter;
我建议您查看您的代码并找到其他将整数转换为char 的地方并相应地进行更改。
顺便说一句，大多数井字游戏使用 ' '、'X' 和 'O' 作为值。
【讨论】：
感谢您的回复，这是个大问题。但我没有使用空格，因为数字使用户能够输入标签来更改板。以上是关于井字游戏使用 C++ 和多维数组 [关闭]的主要内容，如果未能解决你的问题，请参考以下文章