
Ublas : 
Subject: [ublas] Stacking vectors with expression templates
From: Jesse Perla (jesseperla_at_[hidden])
Date: 20090406 21:46:03
In many places in my algorithms, I am stacking small vectors. I need
semantics where I can call functions like the following:
template<int N, int M>
bounded_vector<double, N+M> my_func(const bounded_vector<double, N>&
v1, const bounded_vector<double, N>& v2)
{
bounded_vector<double, 2> v1;
bounded_vector<double, 3> v2;
//.... stuff
return another_func(stack_vector(v1, v2));
}
My stack vector (specialized for the bounded_vector) is as you would expect.
template<class T, int N, int M>
ublas::bounded_vector<T, N+M> stack_vec(const ublas::bounded_vector<T, N>&
v1, const ublas::bounded_vector<T, M>& v2)
{
ublas::bounded_vector<T, N+M> result;
subrange(result, 0, v1.size()) = v1;
subrange(result, v1.size(), v1.size() + v2.size()) = v2;
return result;
}
And a more general version, but which requires a dynamically allocated
temporary:
template<class Vector_T1, class Vector_T2>
ublas::vector<double> stack_vec2(const ublas::vector_expression<Vector_T1>&
v1, const ublas::vector_expression<Vector_T2>& v2)
{
std::size_t s1 = v1().size();
std::size_t s2 = v2().size();
ublas::vector<double> result(s1 + s2);
subrange(result, 0, s1) = v1;
subrange(result, s1, s1 + s2) = v2;
return result;
}
I am terrified that I will have 1 temporary caused inside of the
stack_vector. Another calculating the transformation within another_func
(because it is unlikely to be able to use the return value optimization),
and then a vector copy from the caller of my_func. Obviously I can't get
rid of them all, but it sure would be nice to get rid of some of them.
I never really have figured out the ublas expression templates or there
limitations, but I was wondering if it is possible to return an expression
template to eliminate at least some of this. Any idea of how it would work
or if it is a good idea? Is it hard?
(I am looking for something similar with matrix stacking, but that is
another day).
Thanks,
Jesse