Le 02/11/11 23:56, Mateusz Łoskot a écrit :

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?

Hi, try adding #include <boost/fusion/include/io.hpp> It is at the end of [2] http://www.boost.org/doc/libs/1_47_0/libs/fusion/doc/html/fusion/sequence/op... Best, Vicente