Question

我正在为扑克游戏编写代码，但是我在找出一种从类对象传递值的方法时遇到了问题。在这种情况下，类卡...对象套牌..我想这样做来模拟交易5卡片给玩家。我创建了另一个类型为卡片的对象，名为player1 ..所以在创建了卡片之后......然后将卡片洗牌......我想从对象卡片中抽取5张卡片并将其传递给对象播放器1 ..

然而，当我打印到屏幕以查看是否传递了值时，我只得到了单元格的地址...而不是值。这是代码

code for class deck

[代码]

 class Card
        {
        public:
            int face;
            int suit;

            void toString() {
                cout<<FaceString[face]<<" of "<<SuitString[suit]<<endl;
            }

        };

[/代码]

code for class of deal

[代码]

 class deal
        {
        public:
        Card deck[52];

            void toDisplay(){
                for (int j = 0; j < 5; j++) {  
                cout<<"player card "; deck[j].toString();
                 }
            }   
        };

[\代码]

[代码]

  //creation of both objects
            Card deck[52];
             Card player1[5];

[/代码}

in main


player1.deck; <<==[Error] request for member 'deck' in 'player1', which is of non-class type 'deal [5]'

Answer 1

你的功能签名错误了：

这是你的原型

void deal(const int wDeck[][13], const char *wFace[],
            const char *wSuit[]);

这是你的功能实现

    void deal(const int wDeck[][13], const char *wFace[],
               const char wSuit[]  )
{
     // blah blah
}

你错过了*

Answer 2

原型与功能定义不匹配，试试这个：

void shuffle(int wDeck[][13]);
void deal(const int wDeck[][13], const char *wFace[], const char* wSuit[]);

void shuffle(int wDeck[][13])
{
    int row;
    int column;
    int card;

    for (card = 1; card <= 52; card++){
        do{
            row = rand() % 4;
            column = rand() % 13;
        } while (wDeck[row][column] != 0);

        wDeck[row][column] = card;
    }
}

void deal(const int wDeck[][13], const char *wFace[], const char* wSuit[])
{
    int card;
    int row;
    int column;

    for (card = 1; card <= 52; card++){

        for (row = 0; row <= 3; row++){

            for (column = 0; column <= 12; column++){
                if (wDeck[row][column] == card){
                    printf("%5s of %-8s%c", wFace[column], wSuit[row],
                        card % 2 == 0 ? '\n' : '\t');

                }
            }
        }

    }
}

通过引用传递值但仍然显示地址而不是值

2 个答案: