Hi, I'm still taking early steps with Boost.Fusion. I'm trying to adapt a bunch of struct and template struct for easy I/O operations. I don't understand one thing from the docs. Here we go; 1) Given the example [1] of template struct adoption: namespace demo { template<typename Name, typename Age> struct employee { Name name; Age age; }; } // Any instantiated demo::employee is now a Fusion sequence BOOST_FUSION_ADAPT_TPL_STRUCT( (Name)(Age), (demo::employee) (Name)(Age), (Name, name) (Age, age)) [1] http://www.boost.org/doc/libs/1_47_0/libs/fusion/doc/html/fusion/adapted/ada... 2) The included comment says "demo::employee is now a Fusion sequence". 3) Jumping to I/O and out docs, reading that [2] "The I/O operators: << and >> work generically on all Fusion sequences. " [2] http://www.boost.org/doc/libs/1_47_0/libs/fusion/doc/html/fusion/sequence/op... 4) Means, I should be able to stream Fusion-adopted demo::employee #include <iostream> #include <string> #include <boost/fusion/adapted/struct/adapt_struct.hpp> #include <boost/fusion/include/adapt_struct.hpp> #include <boost/fusion/sequence/io.hpp> #include <boost/fusion/include/io.hpp> // code from above example goes here int main() { demo::employee<std::string, int> e; std::cout << e << std::endl; } 5) Given what I have learned above, I expect the cout << e to work: $ g++ -I/home/mloskot/dev/boost/_svn/trunk boost_fusion.cpp boost_fusion.cpp: In function ‘int main()’: boost_fusion.cpp:28:18: error: no match for ‘operator<<’ in ‘std::cout << e’ ... I assume complete error is not necessary as my question is of general nature. Is my expectation valid? Best regards, -- Mateusz Loskot, http://mateusz.loskot.net Charter Member of OSGeo, http://osgeo.org Member of ACCU, http://accu.org