Using std::vector as view on to raw memory

Question

I'm using a external library which at some point gives me a raw pointer to an array of integers and a size.

Now I'd like to use std::vector to access and modify these values in place, rather than accessing them with raw pointers.

Here is an articifial example that explains the point:

size_t size = 0;
int * data = get_data_from_library(size);   // raw data from library {5,3,2,1,4}, size gets filled in

std::vector<int> v = ????;                  // pseudo vector to be used to access the raw data

std::sort(v.begin(), v.end());              // sort raw data in place

for (int i = 0; i < 5; i++)
{
  std::cout << data[i] << "\n";             // display sorted raw data 
}

Expected output:

1
2
3
4
5

The reason is that I need to apply algorithms from <algorithm> (sorting, swaping elements etc.) on that data.

On the other hand changing the size of that vector would never be changed, so push_back, erase, insert are not required to work on that vector.

I could construct a vector based on the data from the library, use modify that vector and copying the data back to to library, but that would be two complete copies that I'd like to avoid as the data set could be really big.

Answer 1

C++20's std::span

If you are able to use C++20, you could use std::span which is a pointer - length pair that gives the user a view into a contiguous sequence of elements. It is some sort of a std::string_view, and while both std::span and std::string_view are non-owning views, std::string_view is a read-only view.

From the docs:

The class template span describes an object that can refer to a contiguous sequence of objects with the first element of the sequence at position zero. A span can either have a static extent, in which case the number of elements in the sequence is known and encoded in the type, or a dynamic extent.

So the following would work:

#include <span>
#include <iostream>
#include <algorithm>

int main() {
    int data[] = { 5, 3, 2, 1, 4 };
    std::span<int> s{data, 5};

    std::sort(s.begin(), s.end());

    for (auto const i : s) {
        std::cout << i << "\n";
    }

    return 0;
}

Check it out live

Since std::span is basically pointer - length pair, you can use in a following manner too:

size_t size = 0;
int *data = get_data_from_library(size);
std::span<int> s{data, size};

Note: Not all compilers support std::span. Check compiler support here.

UPDATE

If you are not able to use C++20, you could use gsl::span which is basically the base version of the C++ standard's std::span.

C++11 solution

If you are limited to C++11 standard, you can try implementing your own simple span class:

template<typename T>
class span {
   T* ptr_;
   std::size_t len_;

public:
    span(T* ptr, std::size_t len) noexcept
        : ptr_{ptr}, len_{len}
    {}

    T& operator[](int i) noexcept {
        return *ptr_[i];
    }

    T const& operator[](int i) const noexcept {
        return *ptr_[i];
    }

    std::size_t size() const noexcept {
        return len_;
    }

    T* begin() noexcept {
        return ptr_;
    }

    T* end() noexcept {
        return ptr_ + len_;
    }
};

Check out C++11 version live