I believe that I have found a discrepancy between multi_array and its online documentation. In the reference manual at http://boost.org/libs/multi_array/doc/reference.html both the copy constructor and the assignment operator are described as calling the copy constructor of the stored type (element). No mention is made of the assignment operator of element. But in the implementation it appears that both functions call element's assignment operator (via calls to std::copy or boost::detail::multi_array::copy_n). Furthermore the multi_array assignment operator does not call element's copy constructor at all. Here is a simple program to illustrate the behavior:

more ./testMultiArray.cc

#include <boost/multi_array.hpp> #include <iostream> using std::cout; using std::endl; class Foo { public: Foo() { cout << " Foo default constructor" << endl; } Foo& operator=(const Foo& foo) { cout << " Foo operator=" << endl; return *this; } Foo(const Foo& foo) { cout << " Foo copy constructor" << endl; } }; int main() { cout << "Calling multi_array extents constructor" << endl; boost::multi_array<Foo, 1> myFooArray(boost::extents[2]); cout << "\n\nCalling multi_array copy constructor" << endl; boost::multi_array<Foo, 1> myFooArray2(myFooArray); cout << "\n\nCalling multi_array assignment operator" << endl; myFooArray2 = myFooArray; } Here are the runtime results (compiled with gcc 3.4.5 against Boost 1.34.1):

./testMultiArray Calling multi_array extents constructor Foo default constructor Foo copy constructor Foo copy constructor

Calling multi_array copy constructor Foo default constructor Foo copy constructor Foo copy constructor Foo operator= Foo operator= Calling multi_array assignment operator Foo operator= Foo operator= What are the intended requirements on the stored type? Does it need to be Assignable? The above example works since Foo is Assignable, but for other types (such as nested multi_arrays) the multi_array copy constructor will not work. Could someone either fix multi_array or update the documentation? Shawn Pautz