让我的游戏开始工作时遇到问题...对我可能会做什么的任何想法？我已经完成了大部分代码并对其进行了初始化，但由于某种原因，我认为大多数函数都没有正确的点。此外，drawBoard没有void类型。请帮忙
由于
#include <iostream>
using namespace std;
// initialize a 2d board (3,3) with asterisk
void drawBoard(char board[][]);
char check_Winner(char [][]);
int main(){
char board[3][3]={{'*','*','*'},{'*','*','*'},{'*','*','*'}};
bool win = false;
bool in_row = false;
bool in_col = false;
int row;
int column;
// run a loop:
while (!win) {
// Display the content of the board
drawBoard(char board[][]));
cout << "Player 1's turn:";
// Ask player one to chose a location (row, column)
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
// check if move is valid
//check if location has astericks and its between 0 thru 2
bool valid = false;
while (!valid)
{
//check if the move is within range.
if(!(row <= 3 && row >= 1 && column <= 3 && column >= 1)){
cout << "Error: Either the row or column is not within range of the board.\n";
// ask again (error: row in column not within range)
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
}
else if(board[row-1][column-1] != '*')
{
cout << "Error: The (row,column) is already occupied.\n";
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
}else {
valid = true;
}
}
board[row-1][column-1] = 'X';
drawBoard(char board[][]);
// Check if someone won or if there is a tie
check_Winner(char board[][]);
// Ask player two to chose a location (row, column)
cout << "Player 2's turn:";
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
// check if move is valid
//check if location has astericks and its between 0 thru 2
bool valid = false;
while (!valid)
{
//check if the move is within range.
if(!(row <= 3 && row >= 1 && column <= 3 && column >= 1)){
cout << "Error: Either the row or column is not within range of the board.\n";
// ask again (error: row in column not within range)
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
}
else if(board[row-1][column-1] != '*')
{
cout << "Error: The (row,column) is already occupied.\n";
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
}else
{
valid = true;
}
board[row-1][column-1] = 'O';
}
drawBoard(char board[][])
// Check if someone won or if there is a tie
check_Winner(char board[][]);
}
system("pause");
return 0;
}
char check_Winner(char board[][]){
char winner = 'T';
// Checks for horizontal:
for (int i = 0; i < 3; i++)
if (board[i][0] == board[i][1] && board[i][1] == board[i][2])
winner = board[i][0];
// Checks for vertical:
for (int i = 0; i < 3; i++)
if (board[0][i] == board[1][i] && board[1][i] == board[2][i])
winner = board[0][1];
// Checks for diagnol:
if ((board[0][0] == board[1][1] && board[1][1] == board[2][2]) ||
(board[0][2] == board[1][1] && board[1][1] == board[2][0]))
winner = board[1][1];
// checking the result:
switch (winner) {
case 'T': cout << "IT'S A TIE";/* tie */ break;
case 'X': cout << "PLAYER 1 WON!";/* X won */ break;
case 'O': cout << "PLAYER 2 WON!";/* O won */ break;
default : cout << "NEXT PLAYER'S TURN...>"; continue;
}
}
void drawBoard(char board[][])
{
cout << " 1 2 3" << endl;
cout << " +---+---+---+" << endl;
cout << " 1" << " | " << board[0][0] << " | " << board[0][1] << " | " << board[0][2] << " | " << endl;
cout << " +---+---+---+" << endl;
cout << " 2" << " | " << board[1][0] << " | " << board[1][1] << " | " << board[1][2] << " | " << endl;
cout << " +---+---+---+" << endl;
cout << " 3" << " | " << board[2][0] << " | " << board[2][1] << " | " << board[2][2] << " | " << endl;
cout << " +---+---+---+" << endl;
}
答案 0 :(得分：1)
我建议你尝试在你的电路板上使用typedef吗？
typedef char tttboard[3][3];
你的董事会声明：
tttboard board={{'*','*','*'},{'*','*','*'},{'*','*','*'}};
您的功能将如下所示:(通过引用！）
void drawBoard(tttboard& board);
char check_Winner(tttboard& board);
正如您在评论中被告知的那样，您没有char和[][]这样称呼它：
drawBoard(board);
check_Winner(board);
答案 1 :(得分：1)
我将您的代码粘贴到codepad.org并进行编译。第一条错误消息是：
第6行：错误：声明＆＃39; board＆＃39;因为多维数组必须具有除第一个
之外的所有维度的边界
换句话说 - 如果你不告诉编译器某事有关二维数组的维度，它就不会知道如何找到一个元素。这是为什么？好吧，当你访问一个m×n个元素的二维数组时，a [i] [j]，你真的在计算
*(a + n * i + j)
因为每次增加行i时，都会跳过另一个n元素。
另一种思考方式 - ＆＃34; 2D＆＃34;元素真是一大块。如果你有
0 1 2 3
4 5 6 7
8 9 10 11
它们实际上存储为
0 1 2 3 4 5 6 7 8 9 10 11
你会发现元素[1] [1]，比如计算偏移量（1D索引，如果你愿意的话）1 * 4 + 1 = 5。当然，线性数组中的元素5的值为5，这是你可以通过向下和向下找到一个来找到的。
我们通过替换
修复了第一个编译器错误void drawBoard(char board[][]);
char check_Winner(char [][]);
与
void drawBoard(char board[][3]);
char check_Winner(char [][3]);
（顺便说一下，我不确定为什么check_Winner被定义为返回char（你的函数中没有return Winner;语句。 ..），以及为什么你在函数名中使用了下划线和大写字母的奇怪组合。我们稍后会再回过头来看。
接下来，编译器会抱怨您调用这些函数的方式。你需要打电话给他们（在第23行和其他地方）而不是
drawBoard(char board[][]);
但仅仅是
drawBoard(board);
由于您已经定义了传递char，并且您不需要[][]来提醒编译器board的类型（您定义了它）更早）。
编译器抛出的下一个错误（修复后）是
In function 'int main()':
Line 72: error: redeclaration of 'bool valid'
确实，你定义了
bool valid = false;
。之后，当引用相同的变量时，你不能将bool放在变量名前面 - 看起来你是（重新）声明变量。只需加上
valid = true
在第72行。（除此之外 - 你清楚地做了一个大块代码的复制粘贴，这就是这个错误悄然进入的方式。你想要努力想办法制作一个方法来制作一个＆＃34;移动＆ ＃34;可以为玩家1或玩家2调用的单个函数 - 然后你不需要重复所有这些代码，事情看起来会更清晰。而且不容易出错。）。
修复后，编译器会抱怨第99行：
In function 'int main()':
Line 99: error: expected ';' before 'check_Winner'
像这样的错误通常就是这种情况，问题出在前面。第97行：
drawBoard(board)
在结尾处缺少分号，应该是
drawBoard(board);
修复后，编译器会抱怨第130行：
default : cout << "NEXT PLAYER'S TURN...>"; continue;
In function 'char check_Winner(char (*)[3])':
Line 130: error: continue statement not within a loop
continue语句用于表示＆＃34;直接进入您所在循环的下一次迭代＆＃34; （while，for ...）。但是你在switch语句中使用它，并且代码可以跳转到没有循环（在你所在的函数范围内）。如果你只是关闭continue;，代码将继续正常。
事实上（虽然这不是你要问的），break'不会导致程序停止;它只会带你到函数的末尾，代码将继续运行。这就是为什么你真的需要check_Winner函数中的返回值 - 你需要在调用它之后检查它，这样你就可以采取适当的行动（打印消息，并退出游戏）。
在第130行取出continue，我们现在在第133行发出警告：
In function 'char check_Winner(char (*)[3])':
Line 133: warning: control reaches end of non-void function
这就是说＆＃34;你声明这个函数返回char，但是你忘了输入return语句！＆＃34;。添加
return winner;
在功能结束之前，投诉就会消失。
所有这些编辑的结果是以下代码。我并不是说这是好的＆＃34;代码，但至少已删除编译器错误。我希望我所描述的过程对你有所帮助。大多数编译器都有标志来启用警告和错误消息;我强烈建议您始终启用每个警告，并注意编译器抱怨的内容。您将学习编写更好的代码..
免责声明 - 我正在写这篇文章＆＃34;在路上＆＃34;所以我只能使用codepad.org进行基本调试。它不允许交互式编程，所以我无法测试最终的代码是否有效＆＃39; - 我只能告诉你它停止抱怨代码中的错误...我99％确定你的check_Winner代码中有一些逻辑错误 - 一个，你从
winner = 'T';
这意味着除非满足其他条件之一，否则你将打平局;如果你没有留下星号，我认为你应该只打个平手球（如果你真的很聪明，你可以在没有可能的解决方案的时候打个平手球，但这是一个更难以编码的问题）。
无论如何 - 这是编译的代码。在你有一个工作游戏之前还有很多工作要做，我想......
#include <iostream>
using namespace std;
// initialize a 2d board (3,3) with asterisk
void drawBoard(char board[][3]);
char check_Winner(char board[][3]);
int main(){
char board[3][3]={{'*','*','*'},{'*','*','*'},{'*','*','*'}};
bool win = false;
bool in_row = false;
bool in_col = false;
int row;
int column;
// run a loop:
while (!win) {
// Display the content of the board
drawBoard(board);
cout << "Player 1's turn:";
// Ask player one to chose a location (row, column)
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
// check if move is valid
//check if location has astericks and its between 0 thru 2
bool valid = false;
while (!valid)
{
//check if the move is within range.
if(!(row <= 3 && row >= 1 && column <= 3 && column >= 1)){
cout << "Error: Either the row or column is not within range of the board.\n";
// ask again (error: row in column not within range)
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
}
else if(board[row-1][column-1] != '*')
{
cout << "Error: The (row,column) is already occupied.\n";
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
}else {
valid = true;
}
}
board[row-1][column-1] = 'X';
drawBoard(board);
// Check if someone won or if there is a tie
check_Winner(board);
// Ask player two to chose a location (row, column)
cout << "Player 2's turn:";
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
// check if move is valid
//check if location has astericks and its between 0 thru 2
valid = false;
while (!valid)
{
//check if the move is within range.
if(!(row <= 3 && row >= 1 && column <= 3 && column >= 1)){
cout << "Error: Either the row or column is not within range of the board.\n";
// ask again (error: row in column not within range)
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
}
else if(board[row-1][column-1] != '*')
{
cout << "Error: The (row,column) is already occupied.\n";
cout << "What is the row:?\n";
cin >> row;
cout << "What is the column?\n";
cin >> column;
}else
{
valid = true;
}
board[row-1][column-1] = 'O';
}
drawBoard(board);
// Check if someone won or if there is a tie
check_Winner(board);
}
system("pause");
return 0;
}
char check_Winner(char board[][3]){
char winner = 'T';
// Checks for horizontal:
for (int i = 0; i < 3; i++)
if (board[i][0] == board[i][1] && board[i][1] == board[i][2])
winner = board[i][0];
// Checks for vertical:
for (int i = 0; i < 3; i++)
if (board[0][i] == board[1][i] && board[1][i] == board[2][i])
winner = board[0][1];
// Checks for diagnol:
if ((board[0][0] == board[1][1] && board[1][1] == board[2][2]) ||
(board[0][2] == board[1][1] && board[1][1] == board[2][0]))
winner = board[1][1];
// checking the result:
switch (winner) {
case 'T': cout << "IT'S A TIE";/* tie */ break;
case 'X': cout << "PLAYER 1 WON!";/* X won */ break;
case 'O': cout << "PLAYER 2 WON!";/* O won */ break;
default : cout << "NEXT PLAYER'S TURN...>";
}
return winner;
}
void drawBoard(char board[][3])
{
cout << " 1 2 3" << endl;
cout << " +---+---+---+" << endl;
cout << " 1" << " | " << board[0][0] << " | " << board[0][1] << " | " << board[0][2] << " | " << endl;
cout << " +---+---+---+" << endl;
cout << " 2" << " | " << board[1][0] << " | " << board[1][1] << " | " << board[1][2] << " | " << endl;
cout << " +---+---+---+" << endl;
cout << " 3" << " | " << board[2][0] << " | " << board[2][1] << " | " << board[2][2] << " | " << endl;
cout << " +---+---+---+" << endl;
}
后记。
在任何想象中，我都不是C ++专家。但 的人往往坚持不使用using namespace std;，而是写
std::cout << "hello world" << std::endl;
等。随着代码变得越来越复杂，明确哪个函数/常量属于哪个类变得更加重要。所以我被告知......见Why is "using namespace std" considered bad practice?。问题和答案都得到了很多赞成......这表明这是人们想到的重要事情。阅读超出接受的答案......在该页面下面还有一些宝石。