C++ std :: iterator,不在内部使用std :: vector或std :: list

Question

我正在尝试(作为练习)在C++中创建一个简单的数值范围类.它会让你迭代均匀间隔的双精度(如numpy/Python arange):

我想做什么(但使用迭代器):

double lower = ..., upper = ..., delta = ...;
for (double val = lower; val < upper; val += delta)
{
   // do something with val
   f(val);
}
// include the last val to guarantee upper is included or exceeded
f(val); // do something with val

期望的等效迭代器代码:

double lower = ..., upper = ..., delta = ...;
NumericRange nr(lower, upper, delta);
for (NumericRange::const_iterator iter = nr.begin(); iter != nr.end(); iter++)
{
    f(*iter);
}

我希望我的迭代器与STL迭代器兼容,所以我可以重用代码(迭代通过NumericRange应该相当于迭代通过std :: vector).

我只是成功地将值存储在std :: vector中(然后使用std :: vector的迭代器).这就是我在网上发现的一切都解决了这个问题.但是,实际上没有必要存储整个列表.

有没有办法避免存储整个值集？有一些iterableI类可以继承和覆盖++,==等拿到不存储所期望的效果std::vector<double>？

(我真的很想知道如何在没有BOOST的情况下做到这一点,即使它很棒.我问这个是因为我想学习如何写(从头开始)像BOOST解决方案.我当然知道那部分软件工程正在使用其他人创建的工具,但我真的想了解这些工具是如何设计和构建的.)

我的可迭代NumericRange类(在std::vector<double>内部使用):

class NumericRange
{
protected:
  double lower, upper, delta;
  std::vector<double> sorted_range;
public:
  typedef std::vector<double>::const_iterator const_iterator;
  NumericRange()
  {
    lower = upper = delta = std::numeric_limits<double>::quiet_NaN();
    // vector is constructed empty
  }
  NumericRange(double lower_param, double upper_param, double delta_param)
  {
    lower = lower_param;

    upper = upper_param;
    delta = delta_param;
    assert(upper_param > lower_param);

    double val;
    // note: can be much faster without push_back
    for (val = lower_param; val < upper_param; val += delta_param)
      {
    sorted_range.push_back(val);
      }
    // ensure the upper_value is contained or surpassed
    sorted_range.push_back(val);
  }
  // to prevent comparison of the entire vector
  bool operator ==(const NumericRange & rhs) const
  {
    return lower == rhs.lower && upper == rhs.upper && delta == rhs.delta;
  }
  // note: this class doesn't really need to store the values in a
  // vector, but it makes the iterator interface much easier.
  const_iterator begin() const
  {
    return sorted_range.begin();
  }
  const_iterator end() const
  {
    return sorted_range.end();
  }
  double get_lower() const
  {
    return lower;
  }
  double get_upper() const
  {
    return upper;
  }
  double get_delta() const
  {
    return delta;
  }
  size_t size() const
  {
    return sorted_range.size();
  }
  void print() const
  {
    std::cout << "[ " << lower << " : " << upper << ": +=" << delta << " ]" << std::endl;
  }
};

Answer 1

有一些迭代I类可以继承和覆盖++,==等拿到不存储所期望的效果std::vector<double>？

就在这里.它的名字是std::iterator<std::input_iterator_tag, double>.

这是一个开始,使用int.为了节省大脑空间,我使用相同的类来表示范围和迭代器.

#include <iterator>
#include <iostream>

struct NumericRange : public std::iterator< std::input_iterator_tag, int >
{
  int current, fini, delta;
  typedef NumericRange iterator;
  typedef iterator const_iterator;
  iterator begin() { return *this; }
  iterator end() { return iterator(fini, fini, delta); }
  iterator& operator++() { current += delta; return *this; }
  iterator operator++(int) { iterator result(*this); ++*this; return result; }
  int operator*() const { return current; }
  NumericRange(int start, int fini, int delta) 
    : current(start), fini(fini), delta(delta)
  {
  }
  bool operator==(const iterator& rhs) {
    return rhs.current == current;
  }
  bool operator!=(const iterator& rhs) {
    return !(*this == rhs);
  }
};

void f(int i, int j) {
  std::cout << i << " " << j << "\n";
}

int main () {
  int lower = 4, upper = 14, delta = 5;
  NumericRange nr(lower, upper, delta);
  for (NumericRange::const_iterator iter = nr.begin(); iter != nr.end(); iter++)
  {
      f(*iter, *nr.end());
  }
}