Re: [Boost-users] [boost-users] Cartesian product

On 10/26/11 12:12, Christoph Heindl wrote:

On Wed, Oct 26, 2011 at 5:34 PM, Larry Evans wrote:

...

[...] shows how to do this where values in the containers are all same type. It works by simply keeping a vector of loop indices or iterators. During the operator++, it checks which which iterator to increment, and if that is at the end, then it moves to the next iterator. Dereferencing all the iterators should give 1 element in the cross product. I would think that simply changing the vector of loop indices to a fusion vector of those indices would enable the same to be done containers of different types.

Thanks for the hint.

Actually I have something similar up and running based on code from stackoverflow. My problem stems from missing meta-programming skills.

For example if we go for boost fusion, then how to get the value_type of the boost fusion vector corresponding to tuple<....>, etc...

Best regards, Christoph The attached code produces output:

./outer_product_seqs.exe rank=3 ops.size()=18 :i= 0:(type_u<0>=0 type_u<1>=0 type_u<2>=0) :i= 1:(type_u<0>=0 type_u<1>=0 type_u<2>=1) :i= 2:(type_u<0>=0 type_u<1>=0 type_u<2>=2) :i= 3:(type_u<0>=0 type_u<1>=1 type_u<2>=0) :i= 4:(type_u<0>=0 type_u<1>=1 type_u<2>=1) :i= 5:(type_u<0>=0 type_u<1>=1 type_u<2>=2) :i= 6:(type_u<0>=1 type_u<1>=0 type_u<2>=0) :i= 7:(type_u<0>=1 type_u<1>=0 type_u<2>=1) :i= 8:(type_u<0>=1 type_u<1>=0 type_u<2>=2) :i= 9:(type_u<0>=1 type_u<1>=1 type_u<2>=0) :i= 10:(type_u<0>=1 type_u<1>=1 type_u<2>=1) :i= 11:(type_u<0>=1 type_u<1>=1 type_u<2>=2) :i= 12:(type_u<0>=2 type_u<1>=0 type_u<2>=0) :i= 13:(type_u<0>=2 type_u<1>=0 type_u<2>=1) :i= 14:(type_u<0>=2 type_u<1>=0 type_u<2>=2) :i= 15:(type_u<0>=2 type_u<1>=1 type_u<2>=0) :i= 16:(type_u<0>=2 type_u<1>=1 type_u<2>=1) :i= 17:(type_u<0>=2 type_u<1>=1 type_u<2>=2) Compilation finished at Wed Oct 26 16:38:59 Is this what you want? It uses the g++ variadic templates compiler: http://www.generic-programming.org/~dgregor/cpp/variadic-templates.html