Re: Copy-less initialization of a TR1 array

[zr]

On Nov 10, 6:35 pm, Maxim Yegorushkin <maxim.yegorush ...@gmail.com>
wrote:

On Nov 10, 4:20 pm, Vidar Hasfjord <vattilah-gro...@yahoo.co .uk>
wrote:
>

On Nov 10, 12:30 pm, zr <zvir...@gmail. comwrote:

>

Hi,

>

Is there a way to initialize a std::tr1::array with a pre-allocated
built-in array in a copy-less assignment, such that both will point to
the same memory?

>

Yes, you can use 'placement new':

>

    typedef array <int, 4A;
    int init [] = {1, 2, 3, 4};
    A& a = *new (init) A;

>
A may overflow init.
>

Of course, this is equivalent to

>

    array <int, 4a = {1, 2, 3, 4};

>
Only with POD types. If you replace int with a non-trivial type, this
code will end up calling constructors twice and destructors only once.
>
--
Max

I had to write a small test to realize the problem, though i don't
understand what is really happening.
Max, or anyone could you please elaborate?

#include <array>
#include <iostream>

using namespace std::tr1;
using namespace std;

class A
{
public:
A():value(10) { cout << "A() c-tor\n"; }
A(const int& _value):value(_ value) { cout << "A(const int& _a) c-tor
\n"; }
~A() { cout << "A d-tor\n"; }
int value;
};

int main(int argc, char* argv[])
{
A cArray[] = { 1 , 2, 3, 4};

array<A,4>& tr1Array = *new(cArray) array<A,4>;

for (int i=0; i<4; i++)
cout << cArray[i].value << " ";
cout << endl;
for (int i=0; i<4; i++)
cout << tr1Array[i].value << " ";
cout << endl;

return 0;
}

output:
A(const int& _a) c-tor
A(const int& _a) c-tor
A(const int& _a) c-tor
A(const int& _a) c-tor
A() c-tor
A() c-tor
A() c-tor
A() c-tor
10 10 10 10
10 10 10 10
A d-tor
A d-tor
A d-tor
A d-tor