Question

我试图创建一个“战争”的游戏，现在我只是试图让甲板设置，并且我得到错误：

Running /home/ubuntu/workspace/Testing__.cc
/home/ubuntu/workspace/Testing__.cc: In function ‘void showName(Deck**)’:
/home/ubuntu/workspace/Testing__.cc:72:19: error: request for member ‘faceNum’ in ‘cards’, which is of non-class type ‘Deck**’
     int q = cards.faceNum;
                   ^
/home/ubuntu/workspace/Testing__.cc:73:22: error: request for member ‘suit’ in ‘cards’, which is of non-class type ‘Deck**’
     string s = cards.suit;
                      ^
/home/ubuntu/workspace/Testing__.cc: In function ‘int main()’:
/home/ubuntu/workspace/Testing__.cc:94:22: error: cannot convert ‘Deck*’ to ‘Deck**’ for argument ‘1’ to ‘void makeDeck(Deck**)’
   makeDeck(&cards[52]);
                      ^
/home/ubuntu/workspace/Testing__.cc:102:27: error: cannot convert ‘Deck*’ to ‘Deck**’ for argument ‘1’ to ‘void showName(Deck**)’
     showName(&cards[(r-1)]);

这是我到目前为止的代码：

#include <iostream>
#include <cstdlib> //system("PAUSE");
#include <string>
#include <stdio.h>    //NULL
#include <stdlib.h>   // srand, rand 
#include <ctime>     // time 
using namespace std;

class Deck
{
  public:
    int faceNum;
    string suit;

  friend ostream& operator<<(ostream& os, const Deck& cards);
};

ostream& operator<<(ostream& os, const Deck& cards)
{
    os << cards.faceNum << " of " << cards.suit;
    return os;
}


void makeDeck(Deck *cards[52])
{
  //creates the deck, Ace -> King (1-13) of each suit
  int n=1;
  for (int x=0; x<13; x++)
  {
    cards[x]->faceNum = n;
    cards[x]->suit = "Spades";
    n++;
  }
  n=1;
  for (int x=13; x<26; x++)
  {
    cards[x]->faceNum = n;
    cards[x]->suit = "Clubs";
    n++;
  }
  n=1;
  for (int x=26; x<39; x++)
  {
    cards[x]->faceNum = n;
    cards[x]->suit = "Diamonds";
    n++;
  }
  n=1;
  for (int x=39; x<52; x++)
  {
    cards[x]->faceNum = n;
    cards[x]->suit = "Hearts";
    n++;
  }

}

/*
void selectCard()
{
  //each player will use this
}
*/

void showName(Deck *cards[52]) //displays name based on card.faceNum
{
  for(int c=0;c<52;c++)
  {
    int q = cards.faceNum; ///error here
    string s = cards.suit; ///error here
    if(q == 1)
      cout << "Ace of " << s;
    else if(q >= 2 && q <= 10)
      cout << q << " of " << s;
    else if(q == 11)
      cout << "Jack of " << s;
    else if(q == 12)
      cout << "Queen of " << s;
    else if(q == 13)
      cout << "King of " << s;
  }
}

int main()
{
  int r;
  int n=1;

  Deck cards[52];

  makeDeck(&cards[52]); ///error here

  //Show me "this" card.....
  do
  {
    cout << "(99 will exit)\nShow card (1-52): ";
    cin >> r;
    //cout << cards[(r-1)] << endl;
    showName(&cards[(r-1)]); ///error here

  }while (r != 99);


  //welcome!!
  //srand((unsigned)time(NULL)); //rand cast based on time



  //r = rand() % dtype + 1;

  //create deck

  //use selectCard() to randomly pull from pile
}

（使用Cloud9（c9.io））它还处于早期阶段，所以整个程序都不完整，但我应该能够选择一张卡片（1-52）并显示它：

4 of Clubs

我发现（到目前为止）谷歌搜索是我的指针存在问题...我想....我在main（）函数中有这个全部并且它有效，我把它分成了单独的功能，次要代码编辑和错误出现。

我正试图围绕指针，所以任何帮助/建议将不胜感激。我确信这是一个简单的3秒答案，但我真的很难过......

全部谢谢！

Answer 1

错误消息足够清晰。例如，在函数showName中，参数卡的类型为Deck **，它是指向指针的指针。因此，您可能不会使用成员访问运算符。

void showName(Deck *cards[52]) //displays name based on card.faceNum
{
  for(int c=0;c<52;c++)
  {
    int q = cards.faceNum; ///error here
    string s = cards.suit; ///error here

在函数main中，调用函数makeDeck提供类型为Deck *的参数，因为cards[52]是Deck类型数组的元素，索引为52，表达式＆amp; cards [52]是其地址而函数makeDeck的参数类型为Deck **

void makeDeck(Deck *cards[52])

//...

int main()
{
  int r;
  int n=1;

  Deck cards[52];

  makeDeck(&cards[52]); ///error here

所以你所做的就是你得到的。：）

错误：请求'y'中的成员'x'，这是非类型'Class **'

1 个答案: