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In thinking about C++ iterator question , I wrote this sample program: #include <vector> #include <iostream> #include <iterator> #include <algorithm> template <class T> std::ostream& operator<<(std::ostream&os, const std::vector<T>& v) { os<<"("; std::copy(v.begin(), v.end(), std::ostream_iterator<T>(os, ", ")); return os<<")"; } int main() { std::vector<int> v(3); std::vector<std::vector<int> > vv(3, v); std::cout << v << "\n"; // this line works std::cout << vv << "\n"; // this line produces error } I compile this program with gcc and get the typical 100 lines of errors. The relevant part, I believe, is: it.cc:19: instantiated from here /usr/include/c++/4.4/bits/stream_iterator.h:191: error: no match for ‘operator<<’ in ‘ ((std::ostream_iterator >, char, std::char_traits > )this)->std::ostream_iterator >, char, std::char_traits >::_M_stream << __value’ Why does this fail? In my templated operator<< , I try to specify that any vector , regardless of type, is printable. So why doesn't std::vector<std::vector<>> print? EDIT : Using the following code in the template function makes it work #if 0 std::copy(v.begin(), v.end(), std::ostream_iterator<T>(os, ", ")); #else for(typename std::vector<T>::const_iterator it = v.begin(); it != v.end(); it++) { os<<(*it)<<", "; } #endif

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