Reputation: 4716
As far as I can tell in c++ there is no common base class that covers both iterator and reverse_iterator.
The only suggestion I have seen so far is to get around this using templates ( How to write a function that takes an iterator or collection in a generic way? )
However this solution doesn't seem to work for me.
class MyClass
{
template<typename Iter> Iter* generate_iterator(...params...)
{
//returns either a vector::iterator or vector::reverse_iterator
}
template<typename Iter> void do_stuff(Iter *begin, Iter *end)
{
//does stuff between elements specified by begin and end
//I would like this function to remain agnostic of which direction it is working in!
}
void caller()
{
//I would like this function to remain agnostic of which direction it is working in too...
do_stuff(generate_iterator(blah),generate_iterator(foo));
}
};
In this case, generate_iterator() cannot be used as desired because the compiler complains "generate_iterator is not a member of class MyClass" presumably because I haven't specified it (which I can't in practice as caller should be agnostic of the iterator type).
Can anyone help? Thanks in advance!
edit: as Mark B pointed out generate_iterator must return a pointer - now corrected
update: just started using this http://thbecker.net/free_software_utilities/type_erasure_for_cpp_iterators/start_page.html and it seems to work...
Upvotes: 5
Views: 1939
Reputation: 59811
Use boost::variant or boost::any.
boost::variant< reverse_iterator, iterator >
generate_iterator(...) {
if(...) return iterator();
else return reverse_iterator();
}
// user code
boost::variant< reverse_iterator, iterator > v = generate_iterator();
if(reverse_iterator* it = boost::get<reverse_iterator>(v))
...;
else if(...)
...;
Although the variant
is better accessed through a visitor.
The downside is that you need some boiler plate to extract the proper type and is exactly the reason why something like any_iterator
might be a more sensible choice.
Upvotes: 0
Reputation: 5947
By using boost tuple and boost any , your problem can be easily solved. I wrote a example by using boost::any , see below:
#include <boost/any.hpp>
using boost::any_cast;
#define MSG(msg) cout << msg << endl;
boost::any getIterator(std::vector<int>& vec, bool bReverse)
{
if(!bReverse)
return boost::any(vec.begin());
else
return boost::any(vec.rbegin());
}
int main()
{
std::vector<int> myvec;
myvec.push_back(1);
myvec.push_back(2);
myvec.push_back(3);
typedef std::vector<int>::iterator vecIter;
typedef std::vector<int>::reverse_iterator vecRIter;
try
{
boost::any iter = getIterator(myvec, false);
boost::any iter2 = getIterator(myvec, true);
vecIter it1 = any_cast<vecIter>(iter);
vecRIter it2 = any_cast<vecRIter>(iter2);
MSG(*it1);//output 1
MSG(*it2);//output 3
return true;
}
catch(const boost::bad_any_cast &)
{
return false;
}
}
Upvotes: 1
Reputation: 308206
You can create your own iterator class that knows how to go both directions. Encapsulate both types of iterator and internally select whichever one you were initialized with.
Here's a start:
template<typename Container>
class BiIterator
{
public:
BiIterator(Container::iterator i) : m_fwd(i), m_isforward(true) {}
BiIterator(Container::reverse_iterator i) : m_rev(i), m_isforward(false) {}
bool operator==(const BiIterator & left, const BiIterator & right);
Container::value_type & operator*()
{
if (m_isforward)
return *m_fwd;
return *m_rev;
}
const Container::value_type & operator*() const;
BiIterator & operator++()
{
if (m_isforward)
++m_fwd;
else
++m_rev;
return *this;
}
private:
Container::iterator m_fwd;
Container::reverse_iterator m_rev;
bool m_isforward;
};
Upvotes: 3
Reputation: 96241
In C++ you can't write a function that returns two different types. In your template case it will return one or the other depending on the instantiation. You could possibly return a base pointer to a polymorphic iterator but that would cause me to ask what you're really trying to do here. Even the standard containers don't try to do that: They have begin
and rbegin
to distinguish properly. I would suggest having two separate functions that each do the right thing and return one type of iterator or the other as context dictates.
As a side, note that you can't implicitly determine a template instantiation of a type that's only used for the return type of a function.
Upvotes: 2