Does an allocator.construct loop equal std::uninitialized_copy?

Question

In this context T is a certain type and allocator is an allocator object for that type. By default it is std::allocator<T> but this is not necessarily true.

I have a chunk of memory acquired by allocator.allocate(n). I also have a container con of T objects (say, a std::vector<T>). I want to initialize that chunk of memory with the T object(s).

The location of the chunk of memory is stored in T* data.

Are these two code examples always identical?

#include <memory>

// example 1
std::uninitialized_copy(con.begin(), con.end(), data)

// example 2
std::vector<T>::const_iterator in = con.begin();
for (T* out = data; in != con.end(); ++out, ++in) {
    allocator.construct(out, *in);
}

And for these two?

#include <memory>

T val = T(); // could be any T value

// example 3
std::uninitialized_fill(data, data + n, val)

// example 4
for (T* out = data; out != (data + n); ++out) {
    allocator.construct(out, val);
}

Answer 1

According to this explanations They should do the same, as allocator::construct is said to construct the object and std::uninitialized... also constructs the objects. But I do not know, what exactly the standard says and what freedom you have, when implementing your own allocator::construct.

EDIT: Ok, the C++03 standard states in section 20.1.5 §2 table 32, that construct(p,t) should have the same effect as new ((void*)p) T(t) (for any standard compliant allocator, not only std::allocator). And in 20.4.4.1 §1, that uninitialized_copy should have the same effect as

for (; first != last; ++result, ++first)
    new (static_cast<void*>(&*result))
            typename iterator_traits<ForwardIterator>::value_type(*first);

and in 20.4.4.2 §1, that uninitialized_fill has an effect of

for (; first != last; ++first)
    new (static_cast<void*>(&*first))
            typename iterator_traits<ForwardIterator>::value_type(x);

So I think that doesn't leave any room for them to behave differently. So to answer your question: yes, it does.