C语言 如何使用结构体中包含的枚举作为函数的形参

Question

我有一个下面的函数:

int isXOfAKind(card *hand, int x, enum pips pip) {
    //... do something
}

我希望第三个参数是下面结构体中的点:

typedef struct card {
    enum {ACE=1, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN, JACK, QUEEN, KING} pips;
    enum {SPADES, CLUBS, HEARTS, DIAMONDS} suit;
    char cardName[20];
} card;

My header file:

#include <stdio.h>
#include <stdlib.h>
#define DECKSZ 52
#define HAND_SIZE 5
typedef struct card {
    enum pip {ACE=1, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN, JACK, QUEEN, KING} pips;
    enum suit {SPADES, CLUBS, HEARTS, DIAMONDS} suits;
    char cardName[20];
} card;
extern card deck[];
void initDeck(card[]);
void labelCards(card[]);
void shuffleDeck(card[]);
void displayHand(card*);
void arrangeHand(card*);
void swap(card*, card*);
int isFlush(card*);
int isStraight(card*);
int isXOfAKind(card*, int, enum pips);
int isStraightFlush(card*);
int isFullHouse(card*);
int isTwoPair(card*);

源文件:

#include "Poker.h"
int main(void) {
    card deck[DECKSZ];
    card *pDeck = &deck[0];
    initDeck(deck);
    labelCards(deck);
    shuffleDeck(deck);
    displayHand(deck);
    return EXIT_SUCCESS;
}
void initDeck(card deck[]) {
    int counter;
    for (counter = 0; counter < DECKSZ; counter++) {
        deck[counter].pips = (const)((counter % 13) + 1);
        deck[counter].suits = (const)(counter / 13);
    }
}
void labelCards(card deck[]) {
    static const char *pipNames[] = {"Ace","Two","Three","Four","Five","Six","Seven","Eight","Nine","Ten","Jack","Queen","King"};
    static const char *suitNames[] = {"Spades","Hearts","Diamonds","Clubs"};
    int i;
    for (i = 0; i < DECKSZ; i++) {
        sprintf(deck[i].cardName, "%s of %sn", pipNames[i % 13], suitNames[i / 13]);
    }
}
void shuffleDeck(card deck[]) {
    int i, j;
    for (i = 0; i < DECKSZ; i++) {
        j = rand() % DECKSZ;
        swap(&deck[i], &deck[j]);
    }
}
void displayHand(card hand[]) {
    int i;
    for (i = 0; i < HAND_SIZE; i++) {
        printf("%s", hand[i].cardName);
    }
}
void arrangeHand(card *hand) {
    int i, j;
    for (i = HAND_SIZE-1; i >= 0; i--) {
        for (j = 0; j < i; j++) {
            if ((hand+j)->pips > (hand+j+1)->pips)
                swap(hand+j, hand+j+1);
        }
    }
}
void swap(card *c1, card *c2) {
    card temp;
    temp = *c1;
    *c1 = *c2;
    *c2 = temp;
}
int isFlush(card *hand) {
    int i, result = 0;
    for (i = 0; i < HAND_SIZE - 1; i++) {
        if ((hand+i)->suits != (hand+i+1)->suits) {
            result = 1;
            break;
        }
    }
    return result;
}
int isStraight(card *hand) {
    int i, result = 0;
    for (i = 0; i < HAND_SIZE - 1; i++) {
        if ((hand+i)->pips != (hand+i+1)->pips + 1) {
            result = 1;
            break;
        }
    }
    return result;
}
int isXOfAKind(card *hand, int x, enum pips pip) {
    int result = 0;
    return result;
}
int isStraightFlush(card *hand) {
    int result = 0;
    result += isFlush(hand);
    result += isStraight(hand);
    return result;
}
int isFullHouse(card *hand) {
    int result = 0;
    result += isXOfAKind(hand, 3, hand->pips);
    result += isXOfAKind(hand, 2, hand->pips);
    return result;
}
int isTwoPair(card *hand) {
    int result = 0;
    result += isXOfAKind(hand, 2, hand->pips);
    result += isXOfAKind(hand, 2, hand->pips);
    return result;
}

如何使用pips枚举作为函数的参数?我在Ubuntu中使用GCC，在ANSI c中编码。谢谢!

Answer 1

update

你根据最初的答案正确地修复了你的代码，但你只是有一个错别字，一旦你修复了，就会给你干净的代码:

int isXOfAKind(card*, int, enum pips);
                                ^^^^

应:

int isXOfAKind(card*, int, enum pip);
                                ^^^

在这里:

int isXOfAKind(card *hand, int x, enum pips pip) {
                                       ^^^^^^^^

应:

int isXOfAKind(card *hand, int x, enum pip pips) {
                                       ^^^^^^^^  

原始回答

问题是pips在这里是一个变量而不是类型:

enum {ACE=1, TWO, ... } pips;
                        ^^^^

这将创建一个类型pip:

enum pips {ACE=1, TWO, } myPip;
     ^^^^

查看下面的示例:

#include <stdio.h>
typedef struct card {
    enum pips {ACE=1, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN, JACK, QUEEN, KING} myPip;
    enum {SPADES, CLUBS, HEARTS, DIAMONDS} suit;
    char cardName[20];
} card;
void test( enum pips pip) {
   printf( "pip = %dn", pip ) ;
}
int main()
{
  test( TWO ) ;
  test( KING ) ;
}

如果我们去看C99标准草案我们看到6.7.2.3部分Tags段落6说(强调mine):

形式

的类型说明符

struct-or-union identifieropt { struct-declaration-list }

或

 enum identifieropt { enumerator-list }

或

  enum identifieropt { enumerator-list , }

声明结构、联合或枚举类型。列表定义结构内容，联合内容或枚举内容。如果提供了标识符，¹³⁰⁾类型说明符也声明该标识符为该类型的标记。

脚注130说:

如果没有标识符，则该类型在翻译单元内只能由其所属的声明引用。当然，当声明的类型是一个pedef名称时，后续声明可以使用该类型名称来声明具有指定结构、联合或枚举类型的对象。