在让我的游戏工作时遇到问题...对我可能会做什么有什么想法吗？我已经浏览了大部分代码并对其进行了初始化，但由于某种原因，它说大多数函数我没有正确的点。此外，drawBoard没有空隙类型。请帮忙
谢谢
#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;
}
我把你的代码粘贴到 codepad.org 中并编译了它。第一条错误消息是：
第 6 行：错误：将"board"声明为多维数组必须具有除第一个维度之外的所有维度的边界
换句话说 - 如果你不告诉编译器一些关于二维数组维度的信息，它就不知道如何找到一个元素。为什么？好吧，当你访问一个由m x n个元素组成的2D数组a[i][j]时，你实际上是在计算。
*(a + n * i + j)
因为每次递增行i时，都会跳过另一个n元素。
另一种思考方式 - "2D"元素实际上是一个大块。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
你可以找到元素 [1][1]，比如说，通过计算偏移量(如果你愿意的话，一维索引)1 * 4 + 1 = 5.果然，线性数组中的元素 5 的值为5，这是你通过向下和一次穿越会发现的。
我们通过替换来修复第一个编译器错误
void drawBoard(char board[][]);
char check_Winner(char [][]);
跟
void drawBoard(char board[][3]);
char check_Winner(char [][3]);
(顺便说一句，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称中使用下划线和大写字母的奇怪组合。我们稍后会回到这个问题)
接下来，编译器会抱怨您调用这些函数的方式。 您需要调用它们(在第 23 行和其他地方)而不是
drawBoard(char board[][]);
但简单地作为
drawBoard(board);
由于您已经定义了要传递char，并且您不需要[][]来提醒编译器board的类型(您之前定义了它)。
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'
事实上，你定义了
bool valid = false;
在第 33 行。之后，当引用同一变量时，您不得在变量名称前面放置bool- 看起来您正在(重新)声明变量。只要把
valid = true
在第 72 行。(除此之外 - 你显然做了一大块代码的复制粘贴，这就是这个错误是如何潜入的。你想认真考虑找到一种方法，使"make a move"成为可以为玩家 1 或玩家 2 调用的单个函数 - 那么你不需要重复所有这些代码，事情会看起来更干净。而且不太容易出错。
修复后，编译器抱怨第 99 行：
In function 'int main()':
Line 99: error: expected ';' before 'check_Winner'
与此类错误的情况一样，问题出在前面的行中。第 97 行：
drawBoard(board)
末尾缺少分号，应该是
drawBoard(board);
修复后，编译器抱怨第 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"直接进入您所在循环的下一个迭代"(while，for...)。但是您在switch语句中使用它，并且没有代码可以跳转到的循环(在您所在的函数范围内)。 如果您只是关闭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;它只会把你带到函数的末尾，代码会继续。 这就是为什么你真的需要在check_Winner函数中有一个返回值 - 你需要在调用它之后检查它，以便你可以采取适当的操作(打印消息，然后退出游戏)。
去掉 130 行的continue，我们现在在第 133 行遇到警告：
In function 'char check_Winner(char (*)[3])':
Line 133: warning: control reaches end of non-void function
这是在说"你声明了这个函数来返回一个char，但你忘了输入一个return语句！ 加
return winner;
在功能结束之前，投诉就消失了。
所有这些编辑的结果是以下代码。我并不是说它是"好"的代码，但至少编译器错误已被删除。我希望我描述的过程对你的学习有所帮助。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- 我在"在路上"写这篇文章，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代码是否"有效"——我只能告诉你，它不再抱怨代码中的错误......我 99% 确定您的check_Winner代码中有一些逻辑错误 - 首先，您从
winner = 'T';
这意味着除非满足其他条件之一，否则您将拨打平局;我认为只有在没有星号的情况下才应该调用领带(如果你想变得非常聪明，你可以在没有可能的解决方案时调用领带，但这是一个更难编码的问题)。
无论如何 - 这是编译的代码。在你有一个工作游戏之前，还有更多的事情要做，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;
}
附言。
无论如何，我都不是一个C++专家。但是那些倾向于坚持不使用using namespace std;的人，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... 请参阅为什么"使用命名空间 std"被视为不良做法？。问题和答案都得到了很多赞成票......建议这是人们思考的重要事情。请阅读超出接受的答案...在该页面的下方有一些宝石。
我可以建议你尝试在你的电路板上使用typedef吗？
typedef char tttboard[3][3];
您的董事会声明：
tttboard board={{'*','*','*'},{'*','*','*'},{'*','*','*'}};
您的函数将如下所示：(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，您在没有char的情况下称呼它，并像这样[][]：
drawBoard(board);
check_Winner(board);