C++ STL范围容器

Question

我正在寻找一个从双指针映射到对象指针的容器.但是,每个键只是一个与该对象相对应的双精度范围.

例如,可能有一个键值/值对<(0.0 3.0),ptr>或<(3.5 10.0),ptr2>

container [1.0]应该返回ptr,container [3.0]也应该返回ptr,而容器[-1.0]应该是未定义的.

默认情况下是否存在具有类似行为的对象,或者我是否必须自己实现？

编辑

这是我编写的实际代码,可能更容易调试/提供建议.

// Behavior: A range is defined mathematically as (min, max]

class dblRange
{
public:
    double min;
    double max;

    dblRange(double min, double max)
    {
        this->min = min;
        this->max = max;
    };

    dblRange(double val)
    {
        this->min = val;
        this->max = val;
    };

    int compare(const dblRange rhs)
    {
        // 1 if this > rhs
        // 0 if this == rhs
        //-1 if this < rhs
        if (rhs.min == rhs.max && min == max)
        {
            /*if (min > rhs.min)
                return 1;
            else if (min == rhs.min)
                return 0;
            else
                return -1;*/
            throw "You should not be comparing values like this. :(\n";
        }
        else if (rhs.max == rhs.min)
        {
            if (min > rhs.min) 
                return 1;
            else if (min <= rhs.min && max > rhs.min)
                return 0;
            else // (max <= rhs.min)
                return -1;
        }
        else if (min == max)
        {
            if (min >= rhs.max)
                return 1;
            else if (min < rhs.max && min >= rhs.min)
                return 0;
            else // if (min < rhs.min
                return -1;
        }

        // Check if the two ranges are equal:
        if (rhs.min == min && rhs.max == max)
        {
            return 0;
        }
        else if (rhs.min < min && rhs.max <= min)
        {
            // This is what happens if rhs is fully lower than this one.
            return 1;
        }
        else if (rhs.min > min && rhs.min >= max)
        {
            return -1;
        }
        else
        {
            // This means there's an undefined case. Ranges are overlapping, 
            // so comparisons don't work quite nicely.

            throw "Ranges are overlapping weirdly. :(\n";
        }
    };

    int compare(const dblRange rhs) const
    {
        // 1 if this > rhs
        // 0 if this == rhs
        //-1 if this < rhs
        if (rhs.min == rhs.max && min == max)
        {
            /*if (min > rhs.min)
                return 1;
            else if (min == rhs.min)
                return 0;
            else
                return -1;*/
            throw "You should not be comparing values like this. :(\n";
        }
        else if (rhs.max == rhs.min)
        {
            if (min > rhs.min) 
                return 1;
            else if (min <= rhs.min && max > rhs.min)
                return 0;
            else // (max <= rhs.min)
                return -1;
        }
        else if (min == max)
        {
            if (min >= rhs.max)
                return 1;
            else if (min < rhs.max && min >= rhs.min)
                return 0;
            else // if (min < rhs.min
                return -1;
        }

        // Check if the two ranges are equal:
        if (rhs.min == min && rhs.max == max)
        {
            return 0;
        }
        else if (rhs.min < min && rhs.max <= min)
        {
            // This is what happens if rhs is fully lower than this one.
            return 1;
        }
        else if (rhs.min > min && rhs.min >= max)
        {
            return -1;
        }
        else
        {
            // This means there's an undefined case. Ranges are overlapping, 
            // so comparisons don't work quite nicely.

            throw "Ranges are overlapping weirdly. :(\n";
        }
    };

    bool operator== (const dblRange rhs ) {return (*this).compare(rhs)==0;};
    bool operator== (const dblRange rhs ) const {return (*this).compare(rhs)==0;};
    bool operator!= (const dblRange rhs ) {return (*this).compare(rhs)!=0;};
    bool operator!= (const dblRange rhs ) const {return (*this).compare(rhs)!=0;};
    bool operator< (const dblRange rhs ) {return (*this).compare(rhs)<0;};
    bool operator< (const dblRange rhs ) const {return (*this).compare(rhs)<0;};
    bool operator> (const dblRange rhs ) {return (*this).compare(rhs)>0;};
    bool operator> (const dblRange rhs ) const {return (*this).compare(rhs)>0;};
    bool operator<= (const dblRange rhs ) {return (*this).compare(rhs)<=0;};
    bool operator<= (const dblRange rhs ) const {return (*this).compare(rhs)<=0;};
    bool operator>= (const dblRange rhs ) {return (*this).compare(rhs)>=0;};
    bool operator>= (const dblRange rhs ) const {return (*this).compare(rhs)>=0;};

};

现在我很难让地图接受一个双键作为键,即使定义了比较运算符.

这里有一些驱动代码,我用它来测试它是否可行:

std::map<dblRange, int> map;
map[dblRange(0,1)] = 1;
map[dblRange(1,4)] = 2;
map[dblRange(4,5)] = 3;

map[3.0] = 4;

Answer 1

我主要同意Earwicker,因为你可以定义一个范围.现在,我赞成实现具有真正含义的运算符(执行基本类型的操作:如果两个范围相等,则两个范围比较相等).然后,您可以使用第三个map参数将比较函数(或函数)传递给它,以解决此映射的特定问题.

// Generic range, can be parametrized for any type (double, float, int...)
template< typename T >
class range
{
public:
    typedef T value_type;

    range( T const & center ) : min_( center ), max_( center ) {}
    range( T const & min, T const & max )
        : min_( min ), max_( max ) {}
    T min() const { return min_; }
    T max() const { return max_; }
private:
    T min_;
    T max_;
};

// Detection of outside of range to the left (smaller values):
//
// a range lhs is left (smaller) of another range if both lhs.min() and lhs.max() 
// are smaller than rhs.min().
template <typename T>
struct left_of_range : public std::binary_function< range<T>, range<T>, bool >
{
    bool operator()( range<T> const & lhs, range<T> const & rhs ) const
    {
        return lhs.min() < rhs.min()
            && lhs.max() <= rhs.min();
    }
};
int main()
{
    typedef std::map< range<double>, std::string, left_of_range<double> > map_type;

    map_type integer; // integer part of a decimal number:

    integer[ range<double>( 0.0, 1.0 ) ] = "zero";
    integer[ range<double>( 1.0, 2.0 ) ] = "one";
    integer[ range<double>( 2.0, 3.0 ) ] = "two";
    // ...

    std::cout << integer[ range<double>( 0.5 ) ] << std::endl; // zero
    std::cout << integer[ range<double>( 1.0 ) ] << std::endl; // one
    std::cout << integer[ 1.5 ] << std::endl; // one, again, implicit conversion kicks in
}

您必须小心双重值之间的相等和比较.获得相同值(在现实世界中)的不同方式可以产生略微不同的浮点结果.

Answer 2

创建一个类DoubleRange来存储双范围,并在其上实现比较运算符.这样,std::map将以DoubleRange课程为关键,为您完成剩下的工作.