On 4/3/2012 2:45 AM, paul Fultz wrote:
push_back is designed to be non-mutating. That is why. This is true of all the views returned from 'algos', by design. What you can do is to make a container from the view through, e.g. as_vector, as_list. It is *not* true that all sequences return const refs to its elements. Containers definitely return references, otherwise get<N>(c)= expr will not be possible.
Also, front and back return non-const reference, but thats not the point. During iteration, they return const reference. Thats why this is not possible:
struct mutable_func { template<class F> struct result;
template<class F, class T, class U> struct result<F(T, U)> { typedef T type; };
template<class T, class U> T operator()(T && x, U y) const { return x+=y; } };
int i = 0; invoke(foo, tuple<int&, int>(i, 1)); //doesn't compile
here is the output:
What is 'foo'? Anyway, this (below) compiles fine with me with g++ 4.6.1 and MSVC 10: #include <boost/fusion/functional/invocation/invoke.hpp> #include <boost/fusion/tuple.hpp> #include <boost/fusion/sequence/io.hpp> #include <iostream> struct mutable_func { template<class F> struct result; template<class F, class T, class U> struct result<F(T, U)> { typedef T type; }; template<class T, class U> T operator()(T && x, U y) const { return x+=y; } }; int main() { using namespace boost::fusion; int i = 999; std::cout << invoke(mutable_func(), tuple<int&, int>(i, 1)) << std::endl; return 0; } Prints 1000. What am I missing? Regards, -- Joel de Guzman http://www.boostpro.com http://boost-spirit.com