为什么我们不能更改 std::vector 的数据指针？

Question

I have an array char* source and a vector std::vector<char> target . I'd like to make the vector target point to source in O(1), without copying the data.

Something along these lines:

#include <vector>

char* source = new char[3] { 1, 2, 3 };
std::vector<char> target;
target.resize(3);

target.setData(source); // <- Doesn't exist
// OR
std::swap(target.data(), source); // <- swap() does not support char*
delete[] source;

Why is it not possible to manually change where a vector points to? Is there some specific, unmanageable problem that would arise if this was possible?

Answer 1

C++ vector class supports adding and deleting elements, with guaranteed consecutive order in memory. If you could initialize your vector with existing memory buffer, and add enough elements to it, it would either overflow or require reallocation.

The interface of vector assumes that it manages its internal buffer, that is, it can allocate, deallocate, resize it whenever it wants (within spec, of course). If you need something that is not allowed to manage its buffer, you cannot use vector - use a different data structure or write one yourself.

You can create a vector object by copying your data (using a constructor with two pointers or assign ), but this is obviously not what you want.

Alternatively, you can use string_view , which looks almost or maybe exactly what you need.

Answer 2

std::vector is considered to be the owner of the underlying buffer. You can change the buffer but this change causes allocation ie making a copy of the source buffer which you don't want (as stated in the question).

You could do the following:

#include <vector>

int main() {
    char* source = new char[3] { 1, 2, 3 };
    std::vector<char> target;
    target.resize(3);
    target.assign(source, source + 3);
    delete[] source;
    return 0;
}

but againstd::vector::assign :

Replaces the contents with copies of those in the range [first, last).

So copy is performed again. You can't get away from it while using std::vector .

If you don't want to copy data, then you should use std::span from C++20 (or create your own span) or use std::string_view (which looks suitable for you since you have an array of char s).

1st option: Using std::string_view

Since you are limited to C++17, std::string_view might be perfect for you. It constructs a view of the first 3 characters of the character array starting with the element pointed by source .

#include <iostream>
#include <string_view>

int main() {
    char* source = new char[3] { 1, 2, 3 };

    std::string_view strv( source, 3 );

    delete[] source;

    return 0;
}

2nd option: Using std::span from C++20

std::span comes from C++20 so it might not be the most perfect way for you, but you might be interested in what it is and how it works. You can think of std::span as a bit generalized version of std::string_view because it is a contiguous sequence of objects of any type, not just characters. The usage is similar as with the std::string_view :

#include <span>
#include <iostream>

int main() {
    char* source = new char[3] { 1, 2, 3 };

    std::span s( source, 3 );

    delete[] source;

    return 0;
}

3rd option: Your own span

If you are limited to C++17, you can think of creating your own span struct. It might still be an overkill but let me show you (btw take a look at this more elaborated answer ):

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_;
    }
};

int main() {
    char* source = new char[3] { 1, 2, 3 };

    span s( source, 3 );

    delete[] source;

    return 0;
}

So the usage is the same as with the C++20's version of std::span .

Answer 3

The std::string_view and std::span are good things to have (if you have compiler version supporting them). Rolling your own similars is ok too.

But some people miss the whole point why one would want to do this exactly to a vector:

• Because you have an API that gives Struct[] + size_t and give you ownership
• and you also have an API that accepts std::vector<Struct>

ownership could be easily transferred into the vector and no copies made!

You can say: But what about custom allocators, memory mapped file pointers, rom memory that I could then set as the pointer?

• If you are already about to set vector internals you should know what you are doing.
• You can try to supply a "correct" allocator in those cases to your vector actually.

Maybe give a warning on compiling this code yes, but it would be nice if it would be possible.

I would do it this way:

• std::vector would get a constructor that asks for a std::vector::raw_source
• std::vector::raw_source is an uint8_t*, size_t, bool struct (for now)
• bool takeOwnership : tells if we are taking ownership (false => copy)
• size_t size : the size of the raw data
• uint8_t* ptr : the pointer to raw data
• When taking ownership, vector resize and such uses the vectors allocation strategy as you otherwise provided with your template params anyways - nothing new here. If that does not fit the data you are doing wrong stuff!

Yes API I say look more complicated than a single "set_data(..)" or "own_memory(...)" function, but tried to make it clear that anyone who ever uses this api pretty much understands implications of it automagically. I would like this API to exists, but maybe I still overlook other causes of issues?