如何编写一个具有boost :: Range作为参数的函数？

Question

这是实现我自己的函数(例如DoSomethingWithRange)接受增量范围作为参数的好方法吗？

#include <iostream>
#include <vector>
#include <boost/range.hpp>
#include <boost/range/algorithm.hpp>
#include <boost/range/adaptors.hpp>

using namespace std;

template <typename RangeType>
    void DoSomethingWithRange(const RangeType &range)
{
    typename RangeType::const_iterator beginIt = boost::begin(range);
    typename RangeType::const_iterator endIt = boost::end(range);

    for(typename RangeType::const_iterator it = beginIt; it != endIt; ++it)
    {
        cout << *it << endl;
    }
}

bool IsPos(int i)
{
    return i>0;
}

int main(int , char** )
{
    vector<int> test;

    test.push_back(1);
    test.push_back(-1);

    DoSomethingWithRange(test | boost::adaptors::filtered(IsPos));
}

Answer 1

这不适用于普通数组,因为RangeType::const_iterator不会定义.传入时它也不起作用std::pair<iterator,iterator>,Boost.Range也支持它.

相反,你应该使用boost::range_iterator<const RangeType>::type.这将适用于Boost.Range支持的所有类型:普通可迭代对象,数组和迭代器对.

例:

template <typename RangeType>
void DoSomethingWithRange(const RangeType &range)
{
    typedef typename boost::range_iterator<const RangeType>::type const_iterator;
    const_iterator endIt = boost::end(range);
    for(const_iterator it = boost::begin(range); it != endIt; ++it)
        cout << *it << endl;
}

int main(int, char** )
{
    vector<int> test;
    test.push_back(1);
    test.push_back(-1);
    DoSomethingWithRange(test);

    int test2[] = {12,34};
    DoSomethingWithRange(test2);

    std::pair<int*,int*> test3(test2, test2+1);
    DoSomethingWithRange(test3);
}