Re: [Boost-users] generate structs with all combinations of members from a given struct

31 Jan 2010

      On 01/31/10 07:20, Hicham Mouline wrote:
[snip]
...
...
Step1:
  Create cross-product of void and {field1,field2,...}.
  This will produce a sequence of pairs:
no_yes =
     ( (void,field1)
     , (void,field2)
     ...
     , (void,fieldn)
     )
Step2:
Create the crosss product of all rows in no_yes.
cross_no_yes =
    ( (void,void,..void)
    , (field1,void,...void)
    , (void,field2,void...,void)
    ...
    , (field1,field2,...fieldn)
    )
step3:
remove all the voids in all the rows of cross_no_yes.
Since the size == n and each element in that row is size 2,
the cross_no_yes size would be 2^n.  Then simply rm the all void row.
Not tested.
[snip]
My previous attachment could be simplified by using:

   template
   < typename Domains
...
struct cross_product_fold
: fold
   < Domains
   , cross_prod_nil
   , cross_product, arg<2> >
...
{
};
     typedef
   cross_product_fold
   < package
...
::type
absent_fields_seq
;
     typedef
   cross_product_fold
   < absent_fields_seq
::type
select_fields_seq
;
I will try this thanks very much.
I will make all these types derived from a base struct with a virtual dtor.
Then I will move this base type around various translation units (so they
only need to know about the base struct), and then the last translation 
unit
will have access to the same list of types and then dynamic cast the base
struct to each of these generated structs and work with the actual type.
I thought of generation a fusion vector as well but the resulting vector
would look like this:
vector<
 vector<double>,
 vector<double>,
 vector<double>,
 vector<double>,
 vector,
 ...
...
Because all the members of params are double, there is no way to
differentiate between different the 1st and 2nd vector<double>
Yes; however, Couldn't you, pair each type with it's index?
IOW, you have, instead of:

       typedef
     package_c::pkg_type
   fields_t
   ;

from my earlier attachment, you have:

     enum
   field_names
   { field_1
   , field_2
   ...
   , field_n
   };
       typedef
     package
     < pair, vector<double> >
     , pair, vector<double> >
     ...
     , pair, vector >
     ...
     >
   fields_t
   ;

Then, the resulting cross_product would be a sequence of pairs
from which you could generate a fusion map.
...
while with PP, I could name the types   params_field1 and params_field2,
The aforementioned fusion map would have keys like
integral_c, for I=1,N.
You could always write a short template:

   template
     obvious_type_from_map_meta_function?
   get_field(afore_mentioned_map)
   ;

to more conveniently access the named field.

HTH.

Larry

Re: [Boost-users] generate structs with all combinations of members from a given struct

Larry Evans