好的,所以我非常接近完成这个程序。我明白为什么我的计划没有采取行动,而且我能够解决这个问题,但现在我正在努力检查赢家。我意识到我的winGame()
函数应该在某种时候或者在循环结束游戏时做。但是,当我试图做一些调试来分析一些事情时,我意识到一些令人不安的事情。它总是说它是平局,即使它不应该是平局。这些小事是我惭愧不明白的,我真的很想帮助我解决这个问题。另外,我知道如果有一场胜利,应该有一段时间,或者在循环结束时进行。我只是不确定放在哪里,所以如果你有任何建议,请让我知道。 *请注意,在我的有效移动函数中有一个小数组,我计划使它成为一个静态常量数组。我的get函数返回名称中的值(例如getIval()返回单元格对象的初始值),而我的set函数只是适当地分配值。逻辑错误,检查井字游戏的赢家
bool TicTacToe::validMove(char move){
char options[9] = { '1','2', '3', '4','5','6','7', '8','9' };
bool validate = false;
for (int i = 0; i < 9; i++){
if (move == options[i]){
validate = true;
}
}
return (validate);
}
void TicTacToe::setMove(char move){
for (int i = 0; i < ROW; i++){
for (int j = 0; j < COL; j++){
if (board[i][j].getiVal() == move){
board[i][j].setiVal(players[currentPlayer].getMarker());
switchPlayer();
break;
}
}
}
}
void TicTacToe::makeAMove(){
char move;
int turns = 1;
bool validate = true;
do{
cout << "Player " << (getCurrentPlayer() + 1) << " make a move." << endl;
cin >> move;
if (validMove(move)){
if (turns > 4){
cout << "Nested if-else statement." << endl;
winGame();
setMove(move);
}
else
setMove(move);
}
else{
cout << "Invalid Move. Please reenter." << endl;
cin >> move;
}
DrawBoard();
turns++;
} while (turns <= 9);
}
bool TicTacToe::winGame(){
cout << "Calling winGame() " << endl;
bool validate = false;
int k = 0;
for (int i = 0; i < COL; i++){
//check column wins
if (board[0][i].getMarker() == board[1][i].getMarker() && board[1][i].getMarker() == board[2][i].getMarker() && board[2][i].getMarker() != (' ')){
cout << "Column win " << endl;
validate = true;
break;
}
//check row wins
else if (board[i][0].getMarker() == board[i][1].getMarker() && board[i][1].getMarker() == board[i][2].getMarker() && board[i][2].getMarker() != (' ')){
cout << "Row win." << endl;
validate = true;
break;
}
}
if(board[0][0].getMarker() == board[1][1].getMarker() && board[1][1].getMarker() == board[2][2].getMarker() && board[2][2].getMarker() != (' ')){
cout << "Diagonal 1" << endl;
validate = true;
}
else if (board[0][2].getMarker() == board[1][1].getMarker() && board[1][1].getMarker() == board[2][0].getMarker() && board[2][0].getMarker() != (' ')){
cout << "Diagonal 2 " << endl;
validate = true;
}
else{
cout << "It's a draw!" << endl;
validate = true;
}
return (validate);
}
下面是该程序的示例运行供您参考。
//sample run
+--+--+--+
|1 |2 |3 |
+--+--+--+
|4 |5 |6 |
+--+--+--+
|7 |8 |9 |
+--+--+--+
Player 1 make a move.
1
+--+--+--+
|X |2 |3 |
+--+--+--+
|4 |5 |6 |
+--+--+--+
|7 |8 |9 |
+--+--+--+
Player 2 make a move.
2
+--+--+--+
|X |O |3 |
+--+--+--+
|4 |5 |6 |
+--+--+--+
|7 |8 |9 |
+--+--+--+
Player 1 make a move.
3
+--+--+--+
|X |O |X |
+--+--+--+
|4 |5 |6 |
+--+--+--+
|7 |8 |9 |
+--+--+--+
Player 2 make a move.
5
+--+--+--+
|X |O |X |
+--+--+--+
|4 |O |6 |
+--+--+--+
|7 |8 |9 |
+--+--+--+
Player 1 make a move.
+--+--+--+
|X |O |X |
+--+--+--+
|4 |O |6 |
+--+--+--+
|7 |8 |9 |
+--+--+--+
Player 1 make a move.
7
Nested if-else statement.
Calling winGame()
It's a draw!
+--+--+--+
|X |O |X |
+--+--+--+
|4 |O |6 |
+--+--+--+
|X |8 |9 |
+--+--+--+
Player 2 make a move.
8
Nested if-else statement.
Calling winGame()
It's a draw!
+--+--+--+
|X |O |X |
+--+--+--+
|4 |O |6 |
+--+--+--+
|X |O |9 |
+--+--+--+
为什么一旦你发现一个胜利者,你就不会立即返回'true'?相反,您正在检测获胜者,而不是立即返回,您的代码会无缘无故地执行更多检查。 – PaulMcKenzie
我的讲师告诉我,在一个函数中有多个回报是不好的编程习惯。但鉴于多人告诉我同样的事情,我开始认为我的导师给了我不正确的信息。现在我知道了,我会考虑这一点。感谢您的答复! –