我正在开展一个最终项目,需要为纸牌游戏实施链接列表。经销商的手和玩家的手都应该是链接列表以及卡片组。
我遇到的问题是当我尝试创建一个将值(卡片列表末尾的卡片)添加到其中的函数时。将列表传递给它自己,它的尾部不会更新它们的值。我可以使函数返回头指针,但是我将无法记下它的尾部。
我真的很抱歉,因为这对我来说是新鲜的,而且在我生命中的这门课程之前从未编程过。如果我的函数中的某些逻辑看起来不必要地难以阅读,或者直接上升没有意义,请原谅我。
我已经无休止地试着去做这项工作,但我有一种感觉,我做的事情从根本上是错误的
typedef struct card_s {
char suit[20];
int face;
struct card_s *next, *previous;
} card;
card* createHands(card* head, card *cards) {
card *tail = NULL, *temp = NULL, *temp1;
// Go to end of deck
while (cards->next != NULL) {
cards = cards->next;
}
temp = (card *)malloc(sizeof(card));
strcpy(temp->suit, cards->suit);
temp->face = cards->face;
if (head == NULL) { // If the list for the hand doesn't exist, create head
head = temp;
}
else {
tail->next = temp;
}
tail = temp;
tail->next = NULL;
temp1 = cards->previous;
free(cards); // to delete the node added from the deck
cards = temp1;
cards->next = NULL;
while (cards->previous != NULL) {
cards = cards->previous;
}
return head;
}
显然,这只会添加我给它的任何值超过以前的值。它不是任何意义上的连接列表。
我将不胜感激任何帮助!
做了很多关于函数假设的假设
typedef struct card_s {
char suit[20];
int face;
struct card_s *next, *previous;
} card;
您想要手中添加一张卡片
card* AddCard(card* hand, card* draw) {
// check input
if(draw == NULL ) {return NULL;}
// create a copy of the card
card* newCard=malloc(sizeof(card));
if(newCard == NULL) {return hand;} // check malloc result
memcpy(newCard, draw, sizeof(card));
newCard->next=NULL;
newCard->previous=NULL;
if(hand==NULL) {return newCard;} // if the hand was null just return the card
// find where to append the card
card* tail=hand;
while(tail->next != NULL) {tail = tail->next;}
// append the card to the end of the hand
tail->next = newCard;
newCard->previous = tail;
return hand;
}
您的代码中存在几个明显的问题:
while (cards->next != NULL)
假设卡不是NULLtail->next = temp;
但是tail是NULL如果我很好理解createHands的第二个参数是牌组,并且目标是如何提取其最后一张牌以将其添加到经销商/玩家列表并返回它的新头。
请注意,当您提取最后一张卡时,您需要将卡片设为card **
,以便能够将该列表设置为NULL。对于经销商/玩家的列表也可以采用相同的方式,但显然您更愿意返回新的头而不是让第一个参数为card ** head
。
假设必须在头上添加卡,解决方案可以是:
card * createHands(card * head, card ** deck) {
if (deck == NULL)
// no available card
return head;
// Go to end of deck
while ((*deck)->next != NULL) {
deck = &(*deck)->next;
}
card * c = *deck; // the extracted card
*deck = NULL; // remove it from the deck
c->previous = NULL;
if (head != NULL) {
c->next = head;
head->previous = c;
}
return c;
}
一个完整的程序来检查:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct card_s {
char suit[20];
int face;
struct card_s *next, *previous;
} card;
card * createHands(card * head, card ** deck) {
if (deck == NULL)
// no available card
return NULL;
// Go to end of deck
while ((*deck)->next != NULL) {
deck = &(*deck)->next;
}
card * c = *deck; // the extracted card
*deck = NULL; // remove it from the deck
c->previous = NULL;
if (head != NULL) {
c->next = head;
head->previous = c;
}
return c;
}
// check
card * mk(const char * s, card * n)
{
card * c = malloc(sizeof(card));
strcpy(c->suit, s);
c->next = n;
if (n != 0)
n->previous = c;
return c;
}
void pr(const char * who, card * c)
{
printf("%s cards :", who);
while (c != NULL) {
printf(" %s", c->suit);
c = c->next;
}
putchar('\n');
}
int main()
{
card * deck = mk("one", mk("two", mk("three", mk("four", mk("five", NULL)))));
card * dealer = NULL;
card * player = NULL;
pr("deck", deck);
puts("->dealer");
dealer = createHands(dealer, &deck);
pr("deck", deck);
pr("dealer", dealer);
puts("->player");
player = createHands(player, &deck);
pr("deck", deck);
pr("player", player);
puts("->dealer");
dealer = createHands(dealer, &deck);
pr("deck", deck);
pr("dealer", dealer);
puts("->player");
player = createHands(player, &deck);
pr("deck", deck);
pr("player", player);
puts("->dealer");
dealer = createHands(dealer, &deck);
pr("deck", deck);
pr("dealer", dealer);
return 0;
}
编译和执行:
pi@raspberrypi:/tmp $ gcc -pedantic -Wextra -Wall -g l.c
pi@raspberrypi:/tmp $ ./a.out
deck cards : one two three four five
->dealer
deck cards : one two three four
dealer cards : five
->player
deck cards : one two three
player cards : four
->dealer
deck cards : one two
dealer cards : three five
->player
deck cards : one
player cards : two four
->dealer
deck cards :
dealer cards : one three five