因此,我正在创建一个与二十一点非常相似的游戏,但是直到31而不是21为止。这段代码几乎印制了套牌,然后对其进行了洗牌。我应该如何使卡具有int值,以便可以添加它们?例如,我将给用户2张卡,然后显示这2张卡的总和,并询问他是否还需要一张卡。如果用户停下来,那么该轮到经销商了。
#include <iostream>
#include <ctime>
#include<string>
#include<cstdlib>
#include "card.h"
#include<iomanip>
using namespace std;
const int CARDS_PER_DECK = 52;
class deckOfCards {
public:
deckOfCards();
void shuffle();
card dealCard();
void printDeck() const;
private:
card* deck;
int currentCard;
};
//prints the deck
void deckOfCards::printDeck() const {
cout << left;
for (int i = 0; i < CARDS_PER_DECK; i++) {
cout << setw(19) << deck[i].print();
if ((i + 1) % 4 == 0)
cout << endl;
}
}
//the deck
deckOfCards::deckOfCards() {
string faces[] = { "A","2","3","4","5","6","7","8","9","10","J","Q","K" };
string suits[] = { "H","D","C","S" };
deck = new card[CARDS_PER_DECK];
currentCard = 0;
for (int count = 0; count < CARDS_PER_DECK; count++) { //populate the deck in order
deck[count] = card(faces[count % 13], suits[count / 13]);
}
}
//shuffles the cards
void deckOfCards::shuffle() {
currentCard = 0;
for (int first = 0; first < CARDS_PER_DECK; first++) {
int second = (rand() + time(0)) % CARDS_PER_DECK;
card temp = deck[first];
deck[first] = deck[second];
deck[second] = temp;
}
}
card deckOfCards::dealCard() {
if (currentCard > CARDS_PER_DECK)
shuffle();
if (currentCard < CARDS_PER_DECK)
return(deck[currentCard++]);
return(deck[0]);
}
int main()
{
deckOfCards deck;
card currentCard;
deck.printDeck();
deck.shuffle();
cout << endl << endl;
deck.printDeck();
deck.shuffle();
cout << endl << endl;
for (int i = 0; i < 2; i++) {
int sum;
currentCard = deck.dealCard();
cout << currentCard.print() << endl;
}
return 0;
}
Card.h
#ifndef H_card
#define H_card
#include <string>
#include <iostream>
using namespace std;
class card {
public:
card(string cardFace, string cardSuit);
string print() const;
card();
private:
string face;
string suit;
};
card::card()
{
}
card::card(string cardFace, string cardSuit) {
face = cardFace;
suit = cardSuit;
}
string card::print() const {
return (face + " " + suit);
}
#endif
答案 0 :(得分:1)
解决此问题的最简单方法可能是将值存储在卡本身上-但在二十一点中,根据其余卡的值,A可以为1或11,因此您需要能够说明这一点。
因此,您可以在card
上添加一个int value
,对于Ace来说是1,对于特许权使用费是10,对于其他每张卡来说都是面值。
int values[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10 };
(创建卡片时,也可以将值计算为std::max(count + 1, 10)
而不是使用数组。)
现在,您需要一种方法来对当前手牌中的卡进行得分。这是一个通用功能,可以与任何容器的卡片一起使用:
template <typename T>
int score_hand(T const & cards) {
int score = 0;
bool ace = false;
for (card const & c : cards) {
if (c.value == 1 && !ace) {
ace = true;
score += 11;
} else {
score += c.value;
}
}
if (ace && score > 31) {
score -= 10;
}
return score;
}
请注意,甲板不应被声明为card* deck;
。您在这里管理自己的内存,而您的实现违反了rule of three/five。而是使用std::vector<card>
(如std::vector<card> deck;
中一样)。
您几乎永远不需要在C ++中显式使用new
。对动态数组使用类似std::vector
的容器。在需要堆分配对象的地方使用std::unique_ptr
或std::shared_ptr
。