Iterating over odd (even) elements only in a range-based loop

Question

Suppose we have a plain array (or other container which supports range-based loops):

const int N = 8;
int arr[N] = {0, 1, 2, 3, 4, 5, 6, 7};

Using indexes or iterators, we can loop over odd elements and increment the index by two:

for (int i = 0; i < N; i+=2)
{
   std::cout << arr[i] << std::endl;
}

How can I get a similar result by using a range-based loop and avoiding explicit iterators/indexes and iteration skipping? Something like this:

for (const auto& v: odd_only(arr))
{
   std::cout << v << std::endl;
}

What does a simple and elegant solution look like? Does the standard library contain something like this?

Answer 1

There's no support for what you request – but you might write your own even_only and odd_only implementations.

Basic idea is to wrap around the normal iterator of the container in question and do a double increment internally each time we increment once externally:

template <typename C, bool IsOdd>
class even_odd_only
{
    C& c;
public:
    class iterator
    {
    public:
        // all the definitions required for iterator!
        // most if not all might simply be derived from C::iterator...

        // copy/move constructor/assignment as needed

        // core of the wrapper: increment twice internally!
        // just doing += 2 is dangerous, though, we might increment beyond
        // the end iterator (undefined behaviour!)additionally, += 2 only
        // is possible for random access iterators (so we limit usability)
        void operator++() { ++b; if(b != e) ++b; }

        // operator* and operator-> (both return *b), post-increment
        // (defined in terms of pre-increment), etc...
        // comparison: only needs to compare b iterators!

    private:
        C::iterator b;
        C::iterator e; // needed for comparison to avoid incrementing beyond!
        iterator(C::iterator b, C::iterator e) : b(b), e(e) { }
    };
    // const_iterator, too; possibly make a template of above
    // and derive const and non-const iterators from?

    even_odd_only(C& c) : c(c) { }

    iterator begin()
    {
        using std::begin;
        using std::end;
        using std::empty;
        auto b = begin(c);
        // should be self-explanatory:
        // skip first element in odd variant (if there is)
        if constexpr(IsOdd) { if(!empty(c)) { ++b; } }
        return iterator(b, end(c));
    };
    iterator end()
    {
        using std::end;
        return iterator(end(c), end(c));
    }
};

template <typename T>
using even_only = even_odd_base<T, false>;
template <typename T>
using odd_only = even_odd_base<T, true>;

As is, it would work even with non-random-access and even non-bidirectional iterators. But especially for RA-iterators, it's less efficient than the classic loop (due to the intermediate if in operator++).

Defining comparison iterators: always operator== and operator!=, only for random access operators you can additionally have operator[<|>|<=|>=] (→ std::enable_if).

You'll find more details about how to write an iterator here – keep in mind when you encounter, though, that std::iterator itself is deprecated now.

Answer 2

As for what you are currently asking; I do not believe anything exists yet. Now as for iterating over a container by some integer N we can do the following; we can write our own for_each type of function. I've written one below and it works like a gem! You may also want to look into the std::advance function as well for it can be another possible implementation. I was checking that out myself as I was writing this function. However; as for c arrays I'm not sure there is much one can do without a bunch of extra code such as class templates, wrappers, etc. Here is my function.

#include <array>
#include <vector>
#include <iterator>

template<typename Container, typename Function>
void for_each_by_n( Container&& cont, Function f, unsigned increment_by = 1) {
    if ( increment_by == 0 ) return; // must check this for no op

    using std::begin;
    auto it = begin(cont);

    using std::end;
    auto end_it = end(cont);

    while( it != end_it ) {
        f(*it);
        for ( unsigned n = 0; n < increment_by; ++n ) {
            if ( it == end_it ) return;
            ++it;
        }
    }
}

int main() {
    std::array<int,8> arr{ 0,1,2,3,4,5,6,7 };
    std::vector<double> vec{ 1.2, 1.5, 1.9, 2.5, 3.3, 3.7, 4.2, 4.8 };

    auto l = [](auto& v) { std::cout << v << ' '; };

    for_each_by_n(arr, l); std::cout << '\n';
    for_each_by_n(vec, l); std::cout << '\n';

    for_each_by_n(arr, l, 2); std::cout << '\n';
    for_each_by_n(arr, l, 4); std::cout << '\n';

    for_each_by_n(vec, l, 3); std::cout << '\n';
    for_each_by_n(vec, l, 5); std::cout << '\n';

    for_each_by_n(arr, l, 8); std::cout << '\n';
    for_each_by_n(vec, l, 8); std::cout << '\n';

    // sanity check to see if it doesn't go past end.
    for_each_by_n(arr, l, 9); std::cout << '\n';
    for_each_by_n(vec, l, 9); std::cout << '\n';

    return 0;
}

-Output-

 0 1 2 3 4 5 6 7
 1.2 1.5 1.9 2.5 3.3 3.7 4.2 4.8
 0 2 4 6 
 0 4
 1.2 2.5 4.2
 1.2 3.7
 0
 1.2
 0
 1.2

What I like about this example above is that not only can you increment through a loop by some integer N; the above function also takes a function pointer, function object, functor, or lambda and it will perform the required action.

In your case you was trying to loop through your container by 2 for ever odd or every even index and within the loop you were printing the results. Here in my example; I'm printing the results in the form of a lambda that is being passed to this function.

However the only caveat with this particular implementation is that it will always start from index 0. You could easily expand on this by introducing another integer parameter as to an offset of where the iteration will begin; but I'll leave that up to you to do as an exercise.

For the time being we have to settle for what C++11 through C++17 has to offer. In the near future we should have many new and powerful features with the release of C++20.