c - 为什么我收到此格式错误 ‘%c’ 需要类型 ‘int’

Question

为什么我的函数会出现此错误或冲突类型？其中玩家只是一个我只想输出的角色，以显示下一个玩家是哪个玩家

错误

tictac.c:94: warning: format ‘%c’ expects type ‘int’, but argument 2 has type ‘char *’
tictac.c:94: warning: format ‘%c’ expects type ‘int’, but argument 2 has type ‘char *’

代码

void move(char player)
{
        int place;
        printf("player %c, enter placement: \n", &player);
        scanf("%d", &place);

        if (place == 1)
                board[0][0] = player;
        else if (place == 2)
                board[0][1] = player;
        else if (place == 3)
                board[0][2] = player;

        else if (place == 4)
                board[1][0] = player;
        else if (place == 5)
                board[1][1] = player;
        else if (place == 6)
                board[1][2] = player;

        else if (place == 7)
                board[2][0] = player;
        else if (place == 8)
                board[2][1] = player;
        else if (place == 9)
                board[2][2] = player;
}

Answer 1

您不应该在 printf 中使用 &player

改变

printf("player %c, enter placement: \n", &player);

至

printf("player %c, enter placement: \n", player);