C++: subvector from a vector as an argument to a function

Question

I use a function from another library which takes any STL sequential container as an argument. Just for arguments sake, I'll keep it as std::vector .

do_something (std::vector<double>& k)
{
    ...
}

I have a whole vector of data (happens to be of type double ) Is there a way to pass a sub vector preferably via reference and without copying and splitting vector before hand , to this function? It is straight forward via copying (by simply making a copy of subvector , returning the vector from function and putting the respective subvector back into vector ). why?:

1. The reason I do not want to copy is entirely due to speed, I will be calling this function many (and I mean many) times and performance is utmost important.
2. The reason I don;t want to split vector before hand is the file IO operations. The vector being sequential allows fast file write particularly in binary mode

In other words, can I do something like this (in pseudocode exmaple below)?

std::vector<double> myVector{1.2, 5.6, 8.6, 2.2, 1.7, 9.1}; // myVeector of size 6

do_something (&myVector[ -- elementss 2,3,4--]);

Any help appreciated, even if it is not possible and if you know a way to copy back and forth efficiently

Note: If it's any help to SO programmers, my vector will be the same size with the same prelocated subvector which is not equal to from myVector.begin() or myVector.back() . Also this will never change for that one vector

Answer 1

No, it can't be done. That's why standard algorithms accept two (or more) iterators as parameters, rather than a container.

You could also look at the Boost.Range library, which allows for range adaptors including slices.

Depending what's in this function template that you call, you might be able to pass a sliced range in place of the sequence, since the concrete ranges defined in Boost do have begin() and end() member functions. But since it takes a non-const reference to a container, there's a chance that it adds or removes elements, in which case you're out of luck.