来自字符串输入的 C++ 枚举值

Question

在这方面已经有一段时间了，希望得到一些帮助。

我要根据字符串输入使用枚举值 Rank 和 Suit 创建并反对 Card。对于 TWO CLUBS 和 ACE DIAMONDS，该字符串将分别采用“2C、AD”等格式。

我尝试使用我在 StackOverflow 其他地方找到的线性搜索函数，它在那里，但它只返回 -1，因为它似乎没有得到正确的输入类型。

我已经能够使用 char * 数组以另一种方式进行转换，但我这辈子都做不到相反的方式。实现见下文:

卡片.h

#ifndef _card_h
#define _card_h

#include <string>
#include <iostream>

using namespace std;


enum Rank{  TWO, THREE, FOUR, FIVE, SIX, SEVEN,
            EIGHT, NINE, TEN, JACK, QUEEN, KING, ACE};


enum Suit{  CLUBS, DIAMONDS, HEARTS, SPADES};

//Character sets used to convert enums to strings
static const char * RankStrings[] = { "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K", "A",};
static const char * SuitStrings[] = { "C", "D", "H", "S" };

class Card {

public:

    ///Constructors and destructors
    Card();
    ~Card();

    ///Accessors
    Rank getRank();
    Suit getSuit();
    string getSuitString();
    string toString(); ///No longer used, but left for testing
    int linearSearch(const char**, const char*, int);

    ///Mutators
    Card(Rank rank, Suit suit);
    Card(string cardStr);


    ///Operators
    bool operator()(Card*, Card*);
    friend ostream& operator<<(ostream&, Card&);

private:
    Suit suit;
    Rank rank;
};

#endif // _random_h

卡片.cpp

#include "card.h"
#include <string>
#include <stdlib.h>
#include <cstring>
#include <iostream>
#include <sstream>
using namespace std;

/**CONSTRUCTORS*/
///Noarg constructor that sets default card values
Card::Card(){
    suit = CLUBS;
    rank = TWO;
}

Card::~Card(){} //Destructor

/**ACCESSORS*/
Rank Card::getRank(void){
    return rank;
}

Suit Card::getSuit(void){
    return suit;
}

string Card::getSuitString(){
    return SuitStrings[suit];
}

/**From tutorial, was used during testing but has been reimplemented
 * as an overloaded operator<< below
 */
string Card::toString(){

    string cardValues;

    cardValues += RankStrings[rank];
    cardValues += SuitStrings[suit];

    return cardValues;
}

/**MUTATORS*/
///Create a new card with given Rank and Suit
Card::Card(Rank cardRank, Suit cardSuit){
    rank = cardRank;
    suit = cardSuit;
}

int Card::linearSearch (const char **Array, const char *searchKey, int arraySize) {
    for (int i = 0; i < arraySize; ++i) {
        if (strcmp(Array[i], searchKey) == 0)
            return i;
    }

    // We didn't find the searchKey in the Array
    return -1;
}

///Sets card rank and suit based on string input
Card::Card(string cardStr){
    stringstream ss;
    string rankStr;

    char *c = &cardStr.at(0);
    cout << *c << endl; //This prints correct value but doesnt work in method below

    int index = linearSearch(RankStrings, c, 13);
    cout << index << endl;


}



/**OPERATORS*/
///Functor to compare two cards for their value
bool Card::operator()(Card* card1, Card* card2){
    return card1->getRank() > card2->getRank();
}

///Puts a string representation of input card on the output stream
ostream& operator<<(ostream& out, Card& card){
    out << RankStrings[card.rank] << SuitStrings[card.suit];
    return out;
}

Answer 1

问题是您正在获取“2C”中“2”的地址，然后将其用作完整的 C 字符串。然后，当您在 linearSearch 中使用 strcmp 时，它会尝试匹配整个字符串 ("2C") 并失败。

解决此问题的一种方法是将 SuitStrings 或 RankStrings 更改为 char 而不是 char * ，但这也涉及更改您的其他方法。

另一种方法是将您正在使用的字符复制到一个单独的字符串中，这样 strcmp 就可以工作了。应该这样做:

char rankStr[2];
rankStr[0] = cardStr[0];
rankStr[1] = '\0'; // NULL terminated.

int index = linearSearch(RankStrings, rankStr, 13);

另一种方法是将 linearSearch 更改为仅查看两个字符串的第一个字符。