c - c中的井字游戏，试图确定谁赢了

Question

试图确定谁在井字游戏中获胜，我是编程新手。目前让用户在仅输入 1 个 X 或 O 输入后就赢了。输入必须使用 2 个整数、行和列输入。非常感谢任何帮助!

#include <stdio.h>
#include <stdlib.h>

void drawBoard(char board[][3])
{
    int rows, columns;
    for ( rows = 0 ; rows < 3 ; rows++ )
    {
            for ( columns = 0 ; columns < 3 ; columns++ )
            {
               if(board[rows][columns]){
                    printf( "|%c", board[rows][columns] );
                }else{
                   printf("| ");
                }
            }
            printf("|\n");
        }
}


 int main()
{
    char game[3][3]={{0}};
    int totalEntry =0,row,column;
    char value;

    while(totalEntry<=9){
         printf("Please choose x or o: ");
          scanf("%c",&value);
          getchar();
          printf("Enter row number: ");
          scanf("%d",&row);
          getchar();
          printf("Enter Column number: ");
          scanf("%d",&column);
          getchar();
          game[row][column] = value;
          drawBoard(game);

    if((game[0][0] == game[0][1]) && (game[0][1] == game[0][2]) && game[0][0] != 'x')
    if((game[1][0] == game[1][1]) && (game[1][1] == game[1][2]) && game[1][0] != 'x')
    if((game[2][0] == game[2][1]) && (game[2][1] == game[2][2]) && game[2][0] != 'x')
    if((game[0][0] == game[1][0]) && (game[1][0] == game[2][0]) && game[0][0] != 'x')
    if((game[0][1] == game[1][1]) && (game[1][1] == game[2][1]) && game[0][1] != 'x')
    if((game[0][2] == game[1][2]) && (game[1][2] == game[2][2]) && game[0][2] != 'x')
    if((game[0][0] == game[1][1]) && (game[1][1] == game[2][2]) && game[0][0] != 'x')
    if((game[2][0] == game[1][1]) && (game[1][1] == game[0][2]) && game[2][0] != 'x')
        printf("User x has won!");


    if((game[0][0] == game[0][1]) && (game[0][1] == game[0][2]) && game[0][0] != 'o')
    if((game[1][0] == game[1][1]) && (game[1][1] == game[1][2]) && game[1][0] != 'o')
    if((game[2][0] == game[2][1]) && (game[2][1] == game[2][2]) && game[2][0] != 'o')
    if((game[0][0] == game[1][0]) && (game[1][0] == game[2][0]) && game[0][0] != 'o')
    if((game[0][1] == game[1][1]) && (game[1][1] == game[2][1]) && game[0][1] != 'o')
    if((game[0][2] == game[1][2]) && (game[1][2] == game[2][2]) && game[0][2] != 'o')
    if((game[0][0] == game[1][1]) && (game[1][1] == game[2][2]) && game[0][0] != 'o')
    if((game[2][0] == game[1][1]) && (game[1][1] == game[0][2]) && game[2][0] != 'o');
        printf("User o has won!");
        break;


    }




    return 0;
}

Answer 1

我会使用像下面这样的东西，使用循环。请注意，由于棋盘被初始化为零，因此将获胜者分配到仍然为零的棋盘位置意味着即使测试所有相等的通过也找不到获胜者。

//declare this outside the main while-loop:
char winner=0;

for(i=0;i<3 && !winner;i++) {
  if(game[i][0]==game[i][1] && game[i][1]==game[i][2])
    winner=game[i][0]; // across
  else if(game[0][i]==game[1][i] && game[1][i]==game[2][i])
    winner=game[0][i]; // down
}
if( !winner && ( game[0][0]==game[1][1] && game[1][1]==game[2][2] ||
                 game[0][2]==game[1][1] && game[1][1]==game[2][0] ) )
  winner = game[1][1]; // diagonal

if(winner) {
  printf( "Winner is %c!\n",winner );
  break;
}