Converting C uint8_t pointer + size combination to C++ iterators

Question

I have some C code that I want to encapsulate in C++ to make it easier to use.

The C code uses a uint8_t*/size_t pair to reference a piece of memory. Can I convert these to C++ iterators with something like std::begin/std::end? I know these functions don't accept pointers, but perhaps there's some other way. I want to avoid having to copy any data.

What I'm looking for is something like this:

void fn(uint8_t* ptr, size_t size) {
    auto begin = std::begin(...);
    auto end = std::end(...);

    // continue to use begin/end similar to std::vector<uint8_t>::iterator
}

The iterators should be usable with the standard library. Specifically I want to use it with std::copy and the std::vector constructor that takes iterators. I know I have other options to copy memory, but I'm looking for encapsulation in C++ types.

I also tried this, but apparently that is a private constructor. (Makes complete sense to me that I can't construct a vector iterator, but I'm just trying things.)

std::vector<uint8_t>::iterator begin(ptr);

I'd also prefer to avoid having to implement my own iterator types.

Answer 1

auto begin = ptr;
auto end = ptr+size;

will do the trick
(iterators are actually modelled after pointers)

Answer 2

You could simply define the function for example like:)

template <class RandomAccessIterator>
void fn( RandomAccessIterator begin, RandomAccessIterator end ) {

    // continue to use begin/end similar to std::vector<uint8_t>::iterator
}

and call it like

fn( ptr, ptr + size );

Of course you could choice any kind of the iterator that is more appropriate for the function.