Reputation:
I am moving an iterator
backward and forwards through a vector
.
I can check if the iterator ran off the end like so:
++my_iterator;
if ( my_iterator == my_vector.end() )
{
--my_iterator; // if I want to stop the iterator at the end.
my_iterator = my_vector.begin(); // if I want the iterator to wraparound.
}
But how do I check if it ran off the beginning?
Edit: So can I do this?
--my_iterator;
if ( my_iterator == my_vector.rend() )
{
my_iterator = my_vector.begin(); // if I want to stop the iterator at the beginning.
my_iterator = --(my_vector.rbegin()); // if I want the iterator to wraparound.
}
Or do I have to do this?
std::vector< T >::iterator temp_reverse_iterator = reverse_iterator< T >( my_iterator );
++temp_reverse_iterator;
if ( temp_reverse_iterator == my_vector.rend() )
{
my_iterator = my_vector.begin(); // if I want to stop the iterator at the beginning.
my_iterator = --(my_vector.end()); // if I want the iterator to wraparound.
}
else
{
my_iterator = temp_reverse_iterator.base(); // do I need to -- this?
}
And are both of these examples logically sound?
Upvotes: 7
Views: 5387
Reputation: 168716
I wonder if it would be easier for you if you used a Boost Circular Buffer instead of a std::vector
as your underlying data structure.
But, to answer your actual question: You check for wrapping past the beginning by checking to see if the iterator equals v.begin()
.
#include <vector>
#include <cassert>
template <class T> void
Increment( typename std::vector<T>::iterator& it, std::vector<T>& v )
{
assert(v.size() > 0);
++it;
if(it == v.end())
it = v.begin();
}
template <class T> void
Decrement( typename std::vector<T>::iterator& it, std::vector<T>& v )
{
assert(v.size() > 0);
if(it == v.begin())
it = v.end();
--it;
}
int main() {
std::vector<int> v;
v.push_back(0);
v.push_back(1);
std::vector<int>::iterator it;
it = v.begin();
Decrement(it, v);
assert(*it == 1);
Increment(it, v);
assert(*it == 0);
}
Upvotes: 5
Reputation: 39304
You can't really, but you can use a reverse_iterator to cycle backwards through the same vector.
Upvotes: 3