Avoiding copy when writing array to vector - c++

Question

But if you want to prevent the 5000 copy (or more) of some potentially ugly heavy objects. Just for performance and ignore readable code you can do this
#include <iostream>
#include <vector>
int main(int argc, char* argv[])
{
std::vector<int> v;
int* a = (int*) malloc(10*sizeof(int));
//int a[10];
for( int i = 0; i<10;++i )
{
*(a + i) = i;
}
delete v._Myfirst; // ? not sure
v._Myfirst = a;
for( int i = 0; i<10;++i )
{
std::cout << v[i] << std::endl;
}
system("PAUSE");
return 0;
}
simply replace the _Myfirst underlying "array". but be very careful, this will be deleted by the vector.
Does my delete/free of my first fail in some cases?
Does this depend on the allocator?
Is there a swap for that? There is the brand new swap for vectors, but how about from an array?
I had to interface with a code I cannot modify (political reasons), which provides me with arrays of objects and have to make them vector of objects.
The pointer seems the right solution, but it's a lot of memory jumps and I have to perform the loop anyway. I was just wondering if there was some way to avoid the copy and tell the vector, "now this is your underlying array".
I agree that the above is highly ugly and this is why I actually reserve and push in my real code. I just wanted to know if there was a way to avoid the loop.
Please don't answer with a loop and a push back, as obviously that's what my "real" code is already doing. If there is no function to perform this swap can we implement a safe one?

Answer 1

Question 1: does my delete/free of my first fail in some cases ?
Yes. You do not know how the memory for _Myfirst is allocated (apart from it uses the allocator). The standard does not specify that that the default allocator should use malloc to allocate the memory so you do not know if delete will work.
Also you are mixing allocation schemes.
You are allocating with malloc(). But expecting the std::vector<> to allocate with new (because you are calling delete).
Also even if the allocator did use new it would be uisng the array version of new and thus you would need to use the array version of delete to reclaim the memory.
Question 2: does this depends on allocator ?
Yes.
Question 3: is there a swap for that ? there is the brand new swap for vectors, but from an array ?
No.
To prevent expensive copies. Use emplace_back().
std::vector<int> v;
v.reserve(10); // reserve space for 10 objects (min).
for( int i = 0; i<10;++i )
{
v.emplace_back(i); // construct in place in the array (no copy).
}

Answer 2

std::vector<SomeUglyHeavyObject*> vect(5000, NULL);
for (int i=0; i<5000; i++)
{
vect[i] = &arr[i];
}
There, avoids copying and doesn't muck with std::vector internals.

Answer 3

If you don't like the way the vector behaves, don't use a vector. Your code depends on implementation details that you MUST not rely upon.
Yes it's ugly as you said. Make it pretty by using the correct container.
Which container you use depends on what operations you need on the container. It may be something as simple as a C-style array will suit your needs. It may be a vector of pointers, or shared pointers if you want the vector to manage the lifetime of the HeavyObjects.
In response to the need to increase the size of collection (see comments below):
std::vector guarantees that the underlying storage is contiguous, so if you increase the size, it will allocate a new larger buffer and copy the old one into the new (using the copy constructor for the objects if one exists.) Then swap the new buffer into place -- freeing the old one using its own allocator.
This will trash your memory unless the buffer you brute forced into the vector was allocated in a manner consistent with the vector's allocator.
So if your original array is big enough, all you need is a "maxUsed" size_t to tell you how big the collection is and where to put new entries. If it's not big enough, then your technique will either fail horribly, or incur the copy costs you are trying to avoid.