Example:
SomeType bar::foo() const {
SomeType retVal;
for (auto i = 0u; i < 10; ++i) {
retVal = boost::range::join(retVal, m_someDataContainer.equal_range(i));
}
return retVal;
}
Lets say, for simplicity, the m_someDataContainer and the bar class are defined as following:
typedef boost::multi_index_container<
int, bmi::indexed_by<bmi::hashed_unique<bmi::tag<struct someTag>,
bmi::identity<int>>>> Data;
class bar {
public:
SomeType foo() const;
private:
Data m_someDataContainer;
};
The questions are: How do I figure out the return type of foo() and how do I join these ranges without using boost::any_range
EDIT1: Looks like it is quite impossible, calling join in loop on previous joined value makes the result type a nested type of joined_range of joined_range of joined_range ... and so on, which, I guess cannot be deduced easily if at all
Since you have (or can generate) a range of ranges of the same type, you need a flattening range. Using Jesse Good's flatten
from Iterating over a range of ranges:
return flatten(boost::irange(0, 10)
| boost::adaptors::transform(
[this](int i){ return m_someDataContainer.equal_range(i); }));
Unfortunately I think that this will probably leave the iterators dangling, so you should adapt the flatten
there to copy the range into its return value; you can do this with multiple inheritance:
template<typename Cont> using FlatIteratorRange
= boost::iterator_range<flattening_iterator<decltype(std::declval<Cont>().begin())>;
template<typename Cont>
struct FlatRange
: private Cont
, public FlatIteratorRange<Cont>
{
explicit FlatRange(Cont const& c)
: Cont(c)
, FlatIteratorRange<Cont>(
flat_iter(this->Cont::begin(), this->Cont::end()),
flat_iter(this->Cont::end()));
{}
}
template<typename Cont>
auto flatten(Cont const& c) -> FlatRange<Cont>
{
return FlatRange<Cont>(c);
}