Question

目前正试图找到一种方法来为Yahtzee游戏定义C中的一组胜利条件。我也想尝试使用结构来做到这一点。值数组是传递给此函数的指针。我计划检查name.howget是否与实际值匹配，然后允许用户选择他/她可用的那些。但我不知道我编码它的方式是否有效。如何在一个结构中获得Full House的所有组合而不必全部输入？

int win_or_lose(int *array_ptr, int sum_of_dice)
{
    struct Win_Conditions{


    char name[7];
            int used;
            int score;
            int howget;
    };
    //THREE OF A KIND STRUCT
    struct Win_Conditions ThreeKind;
    strcpy(ThreeKind.name, "Three of a Kind");
    ThreeKind.used, false;
    ThreeKind.score, sum_of_dice;
    //second set of values is set to four to that is can accept a four of a kind as a three of a kind.
    ThreeKind.howget, 1,1,1 || 2,2,2 || 3,3,3 || 4,4,4 || 5,5,5 || 6,6,6 || 1,1,1,1 || 2,2,2,2 || 3,3,3,3 || 4,4,4,4 || 5,5,5,5 || 6,6,6,6;

    //FOUR OF A KIND STRUCT
    struct Win_Conditions FourKind;
    strcpy(FourKind.name, "Four of a Kind");
    FourKind.used, false;
    FourKind.score, sum_of_dice;
    FourKind.howget, (1,1,1,1) || (2,2,2,2) || (3,3,3,3) || (4,4,4,4) || (5,5,5,5) || (6,6,6,6);

    struct Win_Conditions FullHouse;
    strcpy(FullHouse.name, "Full House");
    FullHouse.used, false;
    FullHouse.score, 25;
    FullHouse.howget, (1,1 && 1,1,1 )|| (1,1 && 2,2,2) || (1,1 && 3,3,3) || (1,1 && 4,4,4) || (1,1 && 5,5,5) || (1,1 && 6,6,6) ||
        (2,2 && 1,1,1 )|| (2,2 && 2,2,2) || (2,2 && 3,3,3) || (2,2 && 4,4,4) || (2,2 && 5,5,5) || (2,2 && 6,6,6) ..... didn't want to input every combo
}

Answer 1

尝试：

//Assumes diceArray is sorted. Seriously, learn qsort from <stdlib>
int isThreeKind(int diceArray[])
{
  int i;
  for(i=0; i<NUM_DICE-2; i++)
  {
    if(diceArray[i] == diceArray[i+1] && diceArray[i] == diceArray[i+2])
    {
      return TRUE;
    }
  }
  return FALSE;
}

然后在win_or_lose中你打电话

if(isThreeKind(diceArray) == TRUE)
{
  score += sum_of_dice;   //or however yatzhee is played
}

如果你真的想将这类东西组合成模块，你会看到面向对象的编程，你会想要尝试C ++或Java或C＃。

使用虚拟函数howget可以创建一个类，子类可以全部定义自己的howget版本。哪种看起来像你计划如何做到这一点。但对于有人出发来说，这可能会有点沉重。

但我认为你应该从关于C语法的教程开始。每个人都从某处开始。那些逗号......

在Structs C中定义条件语句

1 个答案: