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

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::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. [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::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--