c++ - 使用 std::allocator 和 std::move 防止释放的正确方法

Question

正如标题所说，我想知道这是否是防止在 vector<T> a 中重新分配的正确方法搬家时vector<T> a至 vector<T> b .

vector header :

template<class T, class A = std::allocator<T>>
class vector
{
    typedef typename A::size_type size_type;

    A alloc;
    T* start;
    T* end;

    public:

            explicit vector(size_type n, const T& val = T(), const A& =A());
            vector(vector&&);
            ~vector();

};

构造函数:

注意:分配可以抛出 std::bad_alloc .

template<class T, class A>
vector<T,A>::vector(size_type n, const T& val, const A& a)
    : alloc(a)
{

    start = alloc.allocate(n); //allocate memory for n elements.
    end = start + n;

    for(auto p = start; p!=end; p++)
            alloc.construct(p, val); //construct val at allocated memory.

}

移动构造函数:

问题:这是移动 vector v 的正确方法吗？

template<class T, class A>
vector<T,A>::vector(vector&& v)
    :alloc(v.alloc)             // copy the allocator in v.
{

    start = std::move(v.start); // move the pointer previously returned by allocator. 
    end = std::move(v.end);     // same boundary value.
    v.start = v.end = nullptr;  // nullptr to prevent deallocation when v is destroyed.
}

析构函数:

问题:根据cppreference , allocator::deallocate(p, n)从先前由 allocator 返回的指针释放内存和 n必须等于分配给的元素数。如果不是这种情况怎么办？我的回答是 allocator::deallocate如果指针或元素数量为 n，则不执行任何操作, 不等于之前对 allocator::allocate(n) 的调用.这是真的？

template<class T, class A>
vector<T, A>::~vector()
{
    for(auto p = start; p!=end; p++)
            alloc.destroy(p);           //Destroy objects pointed to by alloc. 

    alloc.deallocate(start, end-start); //Deallocate memory.
}

Answer 1

Is this a proper way to move vector v?

看起来不错，但是 std::move 与指针是多余的。此外，您不会释放 end，因此您不需要将其设置为 null。

What if this is not the case? My answer would be that allocator::deallocate does nothing if the pointer or the number of elements, n, is not equal to previous call to allocator::allocate(n). Is this true?

不，行为未定义。您必须传递相同的 n。