Question

我写了一个小类Dice来模仿真实骰子和Singleton可以继承的模板类Dice的行为。我已经为类operator<<编写了Dice但是编译器在找到它时遇到了问题。我为<<，Dice和Sinlgeton<Dice>重置std::vector<int>运算符，这些运算符是从某些Dice方法返回的，并且它很方便。

我在ubuntu上使用Qt creator 2.5和gcc 4.7。

/home/USER/programming/cpp_yahtzee/main.cpp:12：错误：不匹配 '运营商＆lt;＆lt;'在'std :: operator＆lt;＆lt; ＆gt;（（*＆amp; std :: cout），（（const char *）“hello”））＆lt;＆lt; （安培; 单::实例（）） - ＆GT;骰子:: getLastThrow（）”

这是产生此错误的代码：

std::cout << "hello" << Dice::Instance().getLastThrow();

修改然而，这会输出预期完全没有错误： std::cout << Dice::Instance() 也许这是我的编译器gcc/g++ 4.7的问题（试过gcc/g++ 4.6.3，效果是一样的）？

我的sinlgeton班

template <typename T>
class Singleton
{
public:
    static T& Instance();
    Singleton() {}
private:

    //declare them to prevent copies
    Singleton(Singleton const&);
    void operator=(Singleton const&);

};

template<typename T>
T& Singleton<T>::Instance()
{
    static T _instance;
    return _instance;
}

骰子类：

    class Dice : public Singleton<Dice>
    {
    private:
        std::vector<int> _lastThrow;
    public:
        Dice();
        std::vector<int> generateThrow();
        friend std::ostream& operator<<(std::ostream& os, const Dice& dice);
        friend std::ostream& operator<<(std::ostream& os, const Singleton<Dice>& dice);
        friend std::ostream& operator<<(std::ostream& os, const std::vector<int>& vect);

        //accessor method - returning last throw
        const std::vector<int>& getLastThrow();

        //rethrowing {1,4} - dice #1 and #4
        std::vector<int> Rethrow(const std::vector<int>& objects);
    };

std::ostream& operator<<(std::ostream& os, const Dice& dice)
{
    for (std::vector<int>::const_iterator it = dice._lastThrow.begin();  it != dice._lastThrow.end(); ++it) {
        os << *it;
    }
    return os;
}
std::ostream& operator<<(std::ostream& os, const Singleton<Dice>& dice)
{
    for (std::vector<int>::const_iterator it = dice.Instance().getLastThrow().begin();  it != dice.Instance().getLastThrow().end(); ++it) {
        os << *it;
    }
    return os;

}

std::ostream& operator<<(std::ostream& os, const std::vector<int>& vect)
{
    for (std::vector<int>::const_iterator it = vect.begin();  it != vect.end(); ++it) {
        os << *it;
    }
    return os;
}

std::vector<int> Dice::generateThrow()
{
    static std::vector<int> v(5);

    for (std::vector<int>::iterator it = v.begin();  it != v.end(); ++it) {
        (*it) = rand()%(DICE_MAX)+1;
    }
    _lastThrow = v;
    return v;
}

现在我做不到这样的事情：

std::cout << Dice::Instance().generateThrow();

修改 Ilya Lavrenov的方法正在工作，虽然这不是我想要的，因为这需要创建一个局部变量。我在Singleton类的某个地方遇到了问题。

Answer 1

#include <iostream>
#include <vector>

template <typename T>
class Singleton
{
public:
    static T& Instance();
    Singleton() {}
private:

    //declare them to prevent copies
    Singleton(Singleton const&);
    void operator=(Singleton const&);

};

template<typename T>
T& Singleton<T>::Instance()
{
    static T _instance;
    return _instance;
}

class Dice : public Singleton<Dice>
{
private:
    std::vector<int> _lastThrow;
public:
    Dice()
    {
        for (int i = 0; i < 10; ++i)
            _lastThrow.push_back(i);
    }
    std::vector<int> generateThrow();

    //accessor method - returning last throw
    const std::vector<int>& getLastThrow()
    {
        return _lastThrow;
    }

    //rethrowing {1,4} - dice #1 and #4
    std::vector<int> Rethrow(const std::vector<int>& objects);
};

std::ostream& operator<<(std::ostream& os, const Dice& dice)
{
    for (std::vector<int>::const_iterator it = dice.Instance().getLastThrow().begin();  it != dice.Instance().getLastThrow().end(); ++it) {
        os << *it;
    }
    return os;
}
std::ostream& operator<<(std::ostream& os, const Singleton<Dice>& dice)
{
    for (std::vector<int>::const_iterator it = dice.Instance().getLastThrow().begin();  it != dice.Instance().getLastThrow().end(); ++it) {
        os << *it;
    }
    return os;

}

std::ostream& operator<<(std::ostream& os, const std::vector<int>& vect)
{
    for (std::vector<int>::const_iterator it = vect.begin();  it != vect.end(); ++it) {
        os << *it;
    }
    return os;
}

std::vector<int> Dice::generateThrow()
{
    static std::vector<int> v(5);

    for (std::vector<int>::iterator it = v.begin();  it != v.end(); ++it) {
        (*it) = rand()%(354535)+1;
    }
    _lastThrow = v;
    return v;
}

int main()
{
    Singleton<Dice> a;
    std::cout << a << std::endl;

    return 0;
}

您的代码有些变化，现在编译得很好。和运算符＆lt;＆lt;也很好用

Answer 2

是否与拼写错误有关Dice::Instane - ＆gt; Dice::Instance？

Answer 3

有时IDE intrepret不同于某些'const'运算符，例如eclipse和virtual studio。 jsut从参数'const'

中删除

std::ostream& operator<<(std::ostream& os, std::vector<int>& vect)

我认为它很有用，如果没有，请告诉我您使用的编译器版本和IDE。

具有重载的运算符＆lt;＆lt;编译器找不到它

3 个答案: