Question

所以我试图编写一个更改数组之外的函数，就像将值传递给函数作为引用一样。这是我的代码：

std::array<Card, 52> deck {};
std::array<Card, 52> &deckRef = deck;

initializeDeck(deckRef);
printDeck(deckRef);

和

void initializeDeck(std::array<Card, 52> deck){
int16_t cardNumber{0};
for(int counterSuit{0}; counterSuit < CardSuit::MAX_SUITS; ++counterSuit)
{
    for(int counterRank{0}; counterRank < CardRank::MAX_RANKS; ++counterRank)
        {
            Card card { static_cast<CardRank::CardRank>(counterRank),
                        static_cast<CardSuit::CardSuit>(counterSuit)};

            deck[cardNumber] = card;
            ++cardNumber;

        }
}}

void printDeck(std::array<Card, 52> deck){
for(int16_t counter{0}; counter < 52; ++counter)
    {
        printCard(deck[counter]);
    }}

printDeck打印垃圾但是如果我将printCard（）添加到initializeDeck（），它会打印所有卡片，但它不会存储在main（）中的原始数组中。

我这样做时会收到此错误;

void initializeDeck(std::array<Card, 52>& deck)

undefined reference to `initializeDeck(std::array<Card, 52u>)'|

提前致谢！提前谢谢！

Answer 1

您需要将deck作为参考传递，以便函数外部具有更改。即代码应为

void initializeDeck(std::array<Card, 52>& deck){

Answer 2

主题的轻微扩展，但如果您愿意，可以在编译时生成套牌:(此示例需要c ++ 14，但c ++ 11可以通过更多工作来完成）

#include <iostream>
#include <iterator>
#include <algorithm>
#include <array>

enum class Suit { Spades, Diamonds, Clubs, Hearts };
std::ostream& operator<<(std::ostream& os, Suit s) {
    switch (s) {
        case Suit::Spades: return os << "spades";
        case Suit::Diamonds: return os << "diamonds";
        case Suit::Clubs: return os << "clubs";
        case Suit::Hearts: return os << "hearts";
    }
}

enum class Rank { Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King, Ace };
std::ostream& operator<<(std::ostream& os, Rank r) {
    switch (r) {
        case Rank::Two: return os << "two";
        case Rank::Three: return os << "three";
        case Rank::Four: return os << "four";
        case Rank::Five: return os << "five";
        case Rank::Six: return os << "six";
        case Rank::Seven: return os << "seven";
        case Rank::Eight: return os << "eight";
        case Rank::Nine: return os << "nine";
        case Rank::Ten: return os << "ten";
        case Rank::Jack: return os << "jack";
        case Rank::Queen: return os << "queen";
        case Rank::King: return os << "king";
        case Rank::Ace: return os << "ace";
    }
}

struct Card {
    Suit suit;
    Rank rank;
};

std::ostream& operator<<(std::ostream& os, const Card& c) {
    return os << c.rank << " of " << c.suit;
}

constexpr Card make_card(size_t index)
{
    return Card {
        static_cast<Suit>(index / 13),
        static_cast<Rank>(index % 13)
    };
}


using Deck = std::array<Card, 52>;

template<size_t...Is>
constexpr Deck generate_deck_impl(std::index_sequence<Is...>)
{
    return { make_card(Is)... };
}

constexpr Deck generate_deck()
{
    return generate_deck_impl(std::make_index_sequence<52>());
}


int main()
{
    using namespace std;
    auto deck = generate_deck();
    std::copy(begin(deck), end(deck), ostream_iterator<Card>(cout, ", "));
    cout << endl;

    random_shuffle(begin(deck), end(deck));
    std::copy(begin(deck), end(deck), ostream_iterator<Card>(cout, ", "));
    cout << endl;
}

是否可以传递std :: array的引用？

2 个答案: