c - 如何从函数中的变量获取值并在 C 中的另一个函数中使用它？

Question

我试图在程序中处理一些牌，但我不知道的是，从函数中获取值并将其传递到我的其他函数中，而不使用全局变量。

现在，我被困在我的计划中，无法到达任何地方..

我在下面写了一条评论，指出我的问题所在。希望有人能解决这个问题:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define DIAMONDS 0
#define CLUBS 1
#define HEARTS 2
#define SPADES 3
#define COLOR_SIZE 13
#define NR_OF_SUITS 4
#define DECK_SIZE 52
#define JACK 11
#define QUEEN 12
#define KING 13
#define ACE 1

struct Card
{
    int suit;
    int value;
    int *sum;
};

void player1(struct Card* deck);
void printCards(struct Card *cardDeck);
void swapCards(struct Card *cardA, struct Card *cardB);
void shuffleCards(struct Card *cardDeck);

//==================================================================
int main()
{
    srand(time(NULL));
    //struct Card deck[DECK_SIZE];      //Statically allocated array
    struct Card * deck; //Dynamically allocated array
    int index;
    int suit_index;

    deck = (struct Card *)malloc(sizeof(struct Card) * DECK_SIZE);
    for (suit_index = 0; suit_index < NR_OF_SUITS; suit_index++)    /* Initiera kortleken */
        for (index = 0; index < COLOR_SIZE; index++)
        {
            deck[suit_index*COLOR_SIZE + index].suit = suit_index;
            deck[suit_index*COLOR_SIZE + index].value = index;
        }
    shuffleCards(deck);
    player1(deck);
    system("pause");
    return 0;
}
//==================================================================

void printCards(struct Card *cardDeck)
{
    int sum = 0;
    for (int i = 0; i < 1; i++)
    {
        switch (cardDeck[i].value + 1)
        {
        case ACE: printf("Ace ");
            cardDeck[i].value = 11;
            break;
        case JACK: printf("Jack ");
            cardDeck[i].value = 10;
            break;
        case QUEEN: printf("Queen");
            cardDeck[i].value = 10;
            break;
        case KING: printf("King ");
            cardDeck[i].value = 10;
            break;
        default: printf("%d ", cardDeck[i].value);
            break;
        }

        printf("of ");
        switch (cardDeck[i].suit)
        {
        case DIAMONDS: printf("Diamonds ");
            break;
        case HEARTS: printf("Hearts ");
            break;
        case CLUBS: printf("Clubs ");
            break;
        case SPADES: printf("Spades ");
            break;
        default: printf("Something went wrong!! ");
            break;
        }
        printf("\n");
        sum += cardDeck[i].value;
    }
}

//----------------------------------------------------------------------
void swapCards(struct Card * cardA, struct Card *cardB)
{
    struct Card temp;
    temp = *cardA;
    *cardA = *cardB;
    *cardB = temp;
}

//----------------------------------------------------------------------
void shuffleCards(struct Card *cardDeck)
{

    for (int i = 0; i < DECK_SIZE; i++)
        swapCards(&cardDeck[i], &cardDeck[rand() % 52]);
}

//----------------------------------------------------------------------
void player1(struct Card* deck) {
  int sum = 0;
  printCards(deck);

  // Insert code here
}

我在最后一个函数中的想法是使用上面的变量 sum，并对所有卡片求和，所以如果我调用 printcards(deck) 三次，我每次都想总结一下。

这是怎么做到的？

Answer 1

How do I take a value from a variable in a function and use it in another function in C ?

通常，您可以将其作为指针传递并返回到所需的函数。函数签名应该类似于:

variable_type* function (variable_type* modified_variable);

然后调用如下:

new_value = function(&modified_variable);

其中＆符号&表示将进行更改。

就您的情况而言，您可以通过按值返回来实现，只需将 void printCards(struct Card *cardDeck) 修改为:

int printCards(struct Card *cardDeck) {

    // adding a last line that returns the accumulated sum
    return sum;
}

然后在你的函数中使用它，如下所示:

void player1(struct Card* deck) {
    int sum = 0;
    sum = printCards(deck);
}