如何在TIC TAC TOE中检查获胜者，以及如何不让两个玩家进入同一位置

我现在想用我的代码做2件事。1(检查获胜者2(不要让两个球员都在同一位置进入例如，如果player1已经在板上输入值[0] [0] ='x'，并且player2再次进入董事会[0] [0] ='0'，我如何不让播放器再次在板上输入[0] [0]？

同样，我如何检查获胜者？

我试图通过检查

if(board[0][0]==board[0][1]==board[0][2]

然后获胜者为player1。(Player1按下他的输入后(

但是，它没有用。

我想诚实地说，我仍然没有为下一个问题而获得逻辑。

#include <cstdlib>
#include<iostream>
using namespace std;
int choice;

char board[3][3]={{'1','2','3'},{'4','5','6'},{'7','8','9'}};

void display_board()
{
system("cls");
    cout<<"..............................................Tick Cross Game by Pakistani coder........................................"<<endl;
    cout<<"tttttPlayer1[X]ntttttPlayer2[0]n";
    cout<<"ttttt     |     |     n";
    cout<<"ttttt     |     |     n";
    cout<<"ttttt__"<<board[0][0]<<"___ __"<<board[0][1]<<"___ __"<<board[0][2]<<"n";
    cout<<"ttttt     |     |     n";
    cout<<"ttttt     |     |     n";
    cout<<"ttttt__"<<board[1][0]<<"___ __"<<board[1][1]<<"___ __"<<board[1][2]<<"n";
    cout<<"ttttt     |     |     n";
    cout<<"ttttt     |     |     n";
    cout<<"ttttt___"<<board[2][0]<<"__ __"<<board[2][1]<<"___ __"<<board[2][2]<<"n";
    cout<<"ttttt     |     |     n";
   /// cout<<"ttttt     |     |     n";
}
void player_turn()
{
        cout<<"ntPlayer1[X] turn:";
    cin>>choice;

    switch(choice)
    {
    case 1:
        board[0][0]='X';
        break;
    case 2:
        board[0][1]='X';
        break;
    case 3:
        board[0][2]='X';
        break;
    case 4:
        board[1][0]='X';
        break;
           case 5:
        board[1][1]='X';
        break;
           case 6:
        board[1][2]='X';
        break;
           case 7:
        board[2][0]='X';
        break;
           case 8:
        board[2][1]='X';
        break;
           case 9:
        board[2][2]='X';
        break;
        default:
        cout<<"invalid choice";
        break;
        ///value will go there but need to display the board also
    }
    display_board();
    //checkwin();
    cout<<"player 2 turnn";
    cin>>choice;
    switch(choice)
    {

    case 1:
        board[0][0]='0';
        break;

    case 2:
        board[0][1]='0';
        break;
    case 3:
        board[0][2]='0';
        break;
    case 4:
        board[1][0]='0';
        break;
           case 5:
        board[1][1]='0';
        break;
           case 6:
        board[1][2]='0';
        break;
           case 7:
        board[2][0]='0';
        break;
           case 8:
        board[2][1]='0';
        break;
           case 9:
        board[2][2]='0';
        break;
        default:
        cout<<"invalid choice";
        break;
    }
    display_board();
}
int main()
{
   /*for(int i=0;i<3;i++)
   {
       for(int j=0;j<3;j++)
       {
           cout<<board[i][j];
       }
   }*/
   while(1)
   {
    display_board();
player_turn();

    display_board();
    }
}

预期输出：每当两个播放器在同一数组位置板[0] [0]中输入时，就应该显示第二个插入该位置的玩家的"无效选择"。

同样，每当发生Tictactoe的获胜条件时。即，如果1(连续3行具有相同的值2(连续3个COL具有相同的值3(对角元素的值相同(3(