On 01/31/10 08:12, Larry Evans wrote:
On 01/31/10 07:20, Hicham Mouline wrote: [snip]
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<int,9001,9002,9003>::pkg_type fields_t ;
from my earlier attachment, you have:
enum field_names { field_1 , field_2 ... , field_n }; typedef package < pair<integral_c<field_names,field_1>, vector<double> > , pair<integral_c<field_names,field_2>, vector<double> > ... , pair<integral_c<field_names,field_I>, vector<double,double> > ... > fields_t ;
Then, the resulting cross_product would be a sequence of pairs from which you could generate a fusion map. [snip]
Such a sequence of sequence of fusion::pair's is generated by the attached. Generating a fusion map from a sequence of fusion::pairs has not been done. I didn't see any hint here: http://www.boost.org/doc/libs/1_41_0/libs/fusion/doc/html/fusion/container/g... about how to do that. Output from attached is: --{--output-- absent_fields_seq= ( name=1, type=1) ( name=2, type=2) ( name=3, type=1) size<select_fields_seq>::value=8 select_fields_seq= { absent , absent , absent } { absent , absent , ( name= 3, type= 1) } { absent , ( name= 2, type= 2) , absent } { absent , ( name= 2, type= 2) , ( name= 3, type= 1) } { ( name= 1, type= 1) , absent , absent } { ( name= 1, type= 1) , absent , ( name= 3, type= 1) } { ( name= 1, type= 1) , ( name= 2, type= 2) , absent } { ( name= 1, type= 1) , ( name= 2, type= 2) , ( name= 3, type= 1) } rm_absent_seq= { } { ( name= 3, type= 1) } { ( name= 2, type= 2) } { ( name= 2, type= 2) , ( name= 3, type= 1) } { ( name= 1, type= 1) } { ( name= 1, type= 1) , ( name= 3, type= 1) } { ( name= 1, type= 1) , ( name= 2, type= 2) } { ( name= 1, type= 1) , ( name= 2, type= 2) , ( name= 3, type= 1) } --}--output--