我有2个枚举值,它们的等级和适合度的枚举。我想要一个包含如下数据的文件:


  2C 6D 7小时10秒JC


此信息确定纸牌。

6c是6的俱乐部。

JC是俱乐部的杰克,依此类推。 (大小写无关紧要)。

如何获取输入的角色并将其用于制作新的扑克牌?并且为每个可能的字符串指定大小写似乎是很多代码行。能以某种方式简化吗?

#include "header1card.h"
#include "stdafx.h"
#include <string>
#include <vector>
#include <iostream>
#include <fstream>

using namespace std;

int main() {
  Card card;
}

class Card {
private:
 enum struct ranks {two, three, four, five, six, seven, eight,
             nine, ten, jack, queen, king, ace };
 enum struct suits {diamonds, clubs, hearts, spades };
 suits su; //do what with these two?
 ranks ra;
    };



int fileparser(vector<Card> & readcards, char * filename) {
 ifstream filereading;
 filereading.open(filename, ios::in);

 if (filereading.is_open()) {
   //while read the file here. save what is read to a string.
  char temp;

  while (!filereading.eof()) {
   //how make a card based off of the inputtedr chars?
   char readchar = filereading.get(temp );

   switch (readchar) {
   case '2' :

    break;
   case '3' :

    break;

//and so on to make every card? Seems like a lot of lines of code.
   }
   readcards.pop_back( /*After making the card, put it in vector*/   );

  }
    filereading.close();
 }
 else {
  //throw error, cannot find file.
  cout << "Could not find file " + filename << endl; //error on ' + filename'. Says Expression must have integral or unscoped enum type.
  cerr << "COuld not find file " << endl;
     }
    }

最佳答案

首先,我建议重写您的输入代码,如下所示:

Card temp;
while (filereading >> temp) {
    readcards.push_back(temp); // note: push_back, not pop_back
}


有关为什么不检查eof()的信息,请参见this question

因此,我们只需要为我们的课程写一个operator>>即可:

class Card {
public:
    friend istream& operator>>(istream& is, Card& c) {
        std::string name;
        if (is >> name) {
            if (name.size() == 2) {
                // rank is name[0], suit is name[1]
                // go process
            }
            else {
                // set fail state
                is.setstate(std::ios::failbit);
            }
        }
        return is;
    }
};


那么我们如何获得排名和花色呢?这是您问题的主要部分。对于西装,最简单的事情就是开关:

switch (toupper(name[1])) {
case 'D':
    c.suit = Card::diamonds;
    break;
case 'S':
    // etc.
default:
    // set fail state, maybe log some error too
    is.setstate(std::ios::failbit);
}


对于等级,我们可以通过以下方式简化2, 3, ..., 9部分:

if (name[0] >= '2' && name[0] <= '9') {
    int idx = name[0] - '2'; // now it's an int, with value 0 through 7
    c.rank = Card::two + idx; // because enums are ordered
}
else {
    // fallback to switch
    switch (toupper(name[0])) {
    case 'T':
        c.rank = Card::ten;
        break;
    case 'J':
        // etc
    default:
        // set fail state
        is.setstate(std::ios::failbit);
}

10-01 05:32
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