Dimension-independent loop over boost::multi_array?

Question

Say I've got an N-dimensional boost::multi_array (of type int for simplicity), where N is known at compile time but can vary (i.e. is a non-type template parameter). Let's assume that all dimensions have equal size m.

typedef boost::multi_array<int, N> tDataArray;
boost::array<tDataArray::index, N> shape;
shape.fill(m);
tDataArray A(shape);

Now I would like to loop over all entries in A, e.g. to print them. If N was 2 for example I think I would write something like this

  boost::array<tDataArray::index, 2> index;
  for ( int i = 0; i < m; i++ )
  {
    for ( int j = 0; j < m; j++ )
    {
      index = {{ i, j }};
      cout << A ( index ) << endl;
    }
  }

I've used an index object to access the elements as I think this is more flexible than the []-operator here.

But how could I write this without knowing the number of dimensions N. Is there any built-in way? The documentation of multi_array is not very clear on which types of iterators exist, etc. Or would I have to resort to some custom method with custom pointers, computing indices from the pointers, etc.? If so - any suggestions how such an algorithm could look like?

Answer 1

Ok, based on the Google groups discussion already mentioned in one of the comments and on one of the examples from the library itself, here is a possible solution that lets you iterate over all values in the multi-array in a single loop and offers a way to retrieve the index for each of these elements (in case this is needed for some other stuff, as in my scenario).

#include <iostream>
#include <boost/multi_array.hpp>
#include <boost/array.hpp>

const unsigned short int DIM = 3;
typedef double tValue;
typedef boost::multi_array<tValue,DIM> tArray;
typedef tArray::index tIndex;
typedef boost::array<tIndex, DIM> tIndexArray;

tIndex getIndex(const tArray& m, const tValue* requestedElement, const unsigned short int direction)
{
  int offset = requestedElement - m.origin();
  return(offset / m.strides()[direction] % m.shape()[direction] +  m.index_bases()[direction]); 
}

tIndexArray getIndexArray( const tArray& m, const tValue* requestedElement )
{
  tIndexArray _index;
  for ( unsigned int dir = 0; dir < DIM; dir++ )
  {
    _index[dir] = getIndex( m, requestedElement, dir );
  }

  return _index;
}


int main()
{ 
  double* exampleData = new double[24];
  for ( int i = 0; i < 24; i++ ) { exampleData[i] = i; }

  tArray A( boost::extents[2][3][4] );
  A.assign(exampleData,exampleData+24);

  tValue* p = A.data();
  tIndexArray index;
  for ( int i = 0; i < A.num_elements(); i++ )
  {
    index = getIndexArray( A, p );
    std::cout << index[0] << " " << index[1] << " " << index[2] << " value = " << A(index) << "  check = " << *p << std::endl;
    ++p;
  }

  return 0;
}

The output should be

0 0 0 value = 0 check = 0
0 0 1 value = 1 check = 1
0 0 2 value = 2 check = 2
0 0 3 value = 3 check = 3
0 1 0 value = 4 check = 4
0 1 1 value = 5 check = 5
0 1 2 value = 6 check = 6
0 1 3 value = 7 check = 7
0 2 0 value = 8 check = 8
0 2 1 value = 9 check = 9
0 2 2 value = 10 check = 10
0 2 3 value = 11 check = 11
1 0 0 value = 12 check = 12
1 0 1 value = 13 check = 13
1 0 2 value = 14 check = 14
1 0 3 value = 15 check = 15
1 1 0 value = 16 check = 16
1 1 1 value = 17 check = 17
1 1 2 value = 18 check = 18
1 1 3 value = 19 check = 19
1 2 0 value = 20 check = 20
1 2 1 value = 21 check = 21
1 2 2 value = 22 check = 22
1 2 3 value = 23 check = 23

so the memory layout goes from the outer to the inner indices. Note that the getIndex function relies on the default memory layout provided by boost::multi_array. In case the array base or the storage ordering are changed, this would have to be adjusted.

Answer 2

There is a lack of simple boost multi array examples. So here is a very simple example of how to fill a boost multi array using indexes and how to read all the entries using a single pointer.

typedef boost::multi_array<double, 2> array_type;
typedef array_type::index index;

array_type A(boost::extents[3][2]);

//  ------> x 
// | 0 2 4
// | 1 3 5
// v
// y
double value = 0;
for(index x = 0; x < 3; ++x) 
    for(index y = 0; y < 2; ++y)
        A[x][y] = value++;

double* it  = A.origin();
double* end = A.origin() + A.num_elements();
for(; it != end; ++it){
    std::cout << *it << " ";
}

// -> 0 1 2 3 4 5

Answer 3

If you don't need the index, you can simply do:

  for (unsigned int i = 0; i < A.num_elements(); i++ )
  {
    tValue item = A.data()[i];
    std::cout << item << std::endl;
  }

Answer 4

Based on the answers before I produced this nice overloaded version of the insertion operator for boost::multi_arrays

using namespace std;
using namespace boost::detail::multi_array;

template <typename T , unsigned long K>
ostream &operator<<( ostream &os , const boost::multi_array<T , K> &A )
{
 const T* p = A.data();
 for( boost::multi_array_types::size_type i = A.num_elements() ; i-- ; ++p )
 {
  os << "[ ";
  for( boost::multi_array_types::size_type k = 0 ; k < K ; ) {
   os << ( p - A.origin()  ) / A.strides()[ k ] % A.shape()[ k ]
         +  A.index_bases()[ k ];
   if( ++k < K )
    os << ", ";
    }
  os << " ] = " << *p << endl;
  }

 return os;
 }

It's just a streamlined version of answer 1, except it should work with any type T that has a working operator<<. I tested like

 typedef boost::multi_array<double, 3> array_type;
 typedef array_type::index index;

 index x = 3;
 index y = 2;
 index z = 3;

 array_type A( boost::extents[ x ][ y ][ z ] );

 // Assign values to the elements
 int values = 0;
 for( index i = 0 ; i < x ; ++i ) 
  for( index j = 0 ; j < y ; ++j )
   for( index k = 0 ; k < z ; ++k )
    A[ i ][ j ][ k ] = values++;

 // print the results
 cout << A << endl;

and it seems to work:

[ 0, 0, 0 ] = 0
[ 0, 0, 1 ] = 1
[ 0, 0, 2 ] = 2
[ 0, 1, 0 ] = 3
[ 0, 1, 1 ] = 4
[ 0, 1, 2 ] = 5
[ 1, 0, 0 ] = 6
[ 1, 0, 1 ] = 7
[ 1, 0, 2 ] = 8
[ 1, 1, 0 ] = 9
[ 1, 1, 1 ] = 10
[ 1, 1, 2 ] = 11
[ 2, 0, 0 ] = 12
[ 2, 0, 1 ] = 13
[ 2, 0, 2 ] = 14
[ 2, 1, 0 ] = 15
[ 2, 1, 1 ] = 16
[ 2, 1, 2 ] = 17

Hope this is useful to somebody, and thanks a lot for the original answers: it was very useful to me.