std::array iterator range without template?

Question

With C arrays, it is fairly easy to write code that takes arrays of any size:

void func( T* itBegin, T* itEnd );

void main() {
    T arr1[1];
    func( std::begin(arr1), std::end(arr1) );
    T arr2[2];
    func( std::begin(arr2), std::end(arr2) );
}

How can I do that with std::arrays?

void func( ??? itBegin, ??? itEnd );

void main() {
    std::array<T,1> arr1;
    func( std::begin(arr1), std::end(arr1) );
    std::array<T,2> arr2;
    func( std::begin(arr2), std::end(arr2) );
}

The problem is that, in MSVC 2010, std::array<T,N>::iterator is different for different N. Is this a bug in MSVC 2010? If not, what is the rationale of this design? Yes, I could get pointers from the std::array and pass them instead of iterators, but isn't that unnecessarily ugly?

BTW, boost::array<T,N>::iterator are the same for all N.

void func( T* itBegin, T* itEnd ) is already a templated function, so I cannot se how can you comply with that "without template" requirement. — Gorpik
– Gorpik, Commented Apr 17, 2012 at 10:32
@Gorpik: Well, no, it isn't templated. It just uses a type with the unfortunate name T. See main, which also uses T and cannot be templated. — MSalters
– MSalters, Commented Apr 17, 2012 at 11:27

Andrew Tomazos · Accepted Answer · 2012-04-17 10:23:24Z

4

template <class I>
void func(I begin, I end)
{
    for (auto x = begin; x != end; ++x)
        something_with(*x);
}

Define them genericly as a type parameter, and then just use them as if they were pointers. Anything that behaves pointer-like will compile, things that don't, won't.

Pointer-like things include normal pointers, as well as standard library iterators, and anything else that defines operator=, operator* and operator++.

Doing it this way and as you will only ever use matching pairs of begin/end iterator ranges from the same array<N>, then it doesn't matter if array<N>::iterator is a different type to array<M>::iterator.

edited Apr 17, 2012 at 10:23

answered Apr 17, 2012 at 10:10

Andrew Tomazos

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2 Comments

Arno Over a year ago

I know it is easy with a template. But virtual functions, for example, cannot be templates, and templates have to be defined in the header. There is no technical reason why a template is needed.

Andrew Tomazos Over a year ago

@Arno: Type parameters is the C++ idiom for passing iterators. If you're not comfortable with templates, you are going to be miserable using the standard library.

James Kanze · Accepted Answer · 2012-04-17 10:15:53Z

2

As far as I can tell, the standard doesn't require that different sized std::array have the same type of iterator; having a different type for std::array<int, 1> and std::array<int, 2> seems legal (although one might have some opinions with regards to the quality of the implementation).

If this is a problem, you can either use a C style array, or use pointers:

func( &arr1[0], &arr1[0] + arr1.size() );

Neither solution is ideal, but they're the best I can offer.

answered Apr 17, 2012 at 10:15

James Kanze

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4 Comments

visitor Over a year ago

&*begin(arr1), &*end(arr1) ? Might look bizarre at first glance, but should be a common technique with iterators, smart pointers.

Arno Over a year ago

Well, it only works if the underlying memory block is contiguous and compact, which it is for array and vector, but of course not in general.

UncleBens Over a year ago

@Arno: The longer syntax in the answer also requires that.

James Kanze Over a year ago

@Arno Yes, but his problem is uniquely with std::array. (Of course, if you're going to do this a lot, I'd suggest an overloaded begin and end, which do this for an std::array, and forward to std::begin and std::end for anything else.)

Bo Persson · Accepted Answer · 2012-04-17 10:58:10Z

0

template <class I>
void func(I begin, I end);

edited Apr 17, 2012 at 10:58

Bo Persson

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answered Apr 17, 2012 at 10:11

laxris

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