为什么我在 C 中的井字游戏不能正确检测到获胜者？

Question

我有一个简单的井字游戏程序，应该可以工作，但不能。我只做了一个获胜条件，即使我手动将所有空格设置为 x，似乎也没有满足。我已经尝试在 Visual Studio Code 中对其进行调试，但它并不总是有效，而且可能很烦人，它似乎没有调用该函数，我不确定为什么。

#include <stdio.h>
#include <stdbool.h>

//        l       l
//    0   l   1   l   2
//        l       l
// ***********************
//        l       l
//    3   l   4   l   5
//        l       l
// ***********************
//        l       l
//    6   l   7   l   8
//        l       l


char space[9] = {'p','p','p','p','p','p','p','p','p'};//p is just a placement value
unsigned int currentPlayer = 1;
bool winner = false;
int p1;
int p2;

int win();

int win(){
    if(space[0] == 'x' && space[1] == 'x' && space[2] == 'x'){
        printf("Player 1 wins");
        winner = true;
    }
}


int main(){
    while(winner == false){//winenr becomes true when someone wins

        
        //The Board//
        printf("       l       l       \n   %c   l   %c   l   %c   \n       l       l       \n***********************\n       l       l       \n   %c   l   %c   l   %c   \n       l       l       \n***********************\n       l       l       \n   %c   l   %c   l   %c   \n       l       l       \n\n",space[0], space[1], space[2], space[3], space[4], space[5], space[6], space[7], space[8]);
        
        if(winner == false){
        //Player 1's turn//     
            while(currentPlayer == 1 && winner == false){//to keep the loop going if they make a wrong move

                win();//to refer back to the win function to see if win conditions are met

                printf("Player 1:");
                scanf("%d", &p1);
                printf("\n\n\n\n");
            

                if(space[p1] == 'x' || space[p1] == 'o'){//if the space is equal to either letter that means they need to input a new move
                
                    printf("That space is currently taken pick another space\n\n");
                }
                if(space[p1] != 'x' && space[p1] != 'o'){//if neither space is equal to a letter then you can put your move there
                    space[p1] = 'x';
                    currentPlayer = 2;
                    break;
                
                }
                else{ 
                    printf("invalid input\n\n");
                }
            

            } 


            while(currentPlayer == 2 && winner == false){//player 2's turn

                printf("Player 2:");
                scanf("%d", &p2);
                printf("\n\n\n\n");

                if(space[p2] == 'x' || space[p2] == 'o'){
                
                    printf("That space is currently taken pick another space\n\n");
                }
                if(space[p2] != 'x' || space[p2] != 'o'){
                    space[p2] = 'o';
                    currentPlayer = 1;
                    break; //changing the player
                }
                else{ 
                    printf("invalid input\n\n");
                }
            }
        }
        else{
            break;
        }
    }
}

Answer 1

当您的“获胜”条件得到满足时，它正在打印消息 - 但是您不会终止程序。它只是继续。在win() 中将winner = true; 替换为exit(0)。您还需要调用win()在一块被添加到棋盘之前，而不是之前。并对玩家 1 和玩家 2 执行此操作。