> Why not add a helper method to define a defacto > type for it to HDF5 just as many of these other users are currently > doing ABOVE HDF5? That way, they all don't wind up essentially > duplicating the same effort?
To be fair, because there's no one-size-fits-all complex number solution outside Fortran, a one-size-fits-all cross-language HDF5 helper method to define a complex type would be difficult to get right. Not technically difficult, just socially difficult. - Rhys _______________________________________________ Hdf-forum is for HDF software users discussion. [email protected] http://mail.hdfgroup.org/mailman/listinfo/hdf-forum_hdfgroup.org
