I'm trying to understand the rules for alignment with placement new.

If we have

struct A
{
size_t length;
};

my understanding (please correct me if I'm wrong) is that this results
in a
data structure that is properly aligned:

typedef typename std::aligned_storage<sizeof(A),
alignof(A)>::type storage_type;

char* storage = new char [sizeof(storage_type)];
A* pa = new(storage)A();

But what if we have

struct B
{
size_t length;
int *p;
};

and wish to allocate storage for B and p with placement new from the
same storage?

Does this satisfy alignment rules?

typedef typename std::aligned_storage<sizeof(B),
alignof(B)>::type storage_type;

size_t length = 10;
char* storage = new char
[sizeof(storage_type)+length*sizeof(int)];
B* pb = new(storage)B();

pb->p = new(storage + sizeof(storage_type))int[length];

Thanks,
Daniel


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{ edited by mod to shorten lines to ~70 characters. -mod }

On Sunday, 13 December 2015 19:10:14 UTC+1, Daniel wrote:

I'm trying to understand the rules for alignment with placement new.

If we have

struct A
{
size_t length;
};

my understanding (please correct me if I'm wrong) is that this results
in a
data structure that is properly aligned:

typedef typename std::aligned_storage<sizeof(A),
alignof(A)>::type storage_type;

char* storage = new char [sizeof(storage_type)];
A* pa = new(storage)A();

But what if we have

struct B
{
size_t length;
int *p;
};

and wish to allocate storage for B and p with placement new from the
same storage?

Does this satisfy alignment rules?

typedef typename std::aligned_storage<sizeof(B),
alignof(B)>::type storage_type;

size_t length = 10;
char* storage = new char
[sizeof(storage_type)+length*sizeof(int)];
B* pb = new(storage)B();

pb->p = new(storage + sizeof(storage_type))int[length];

I don't think it does. Imagine a crazy system where size_t was unsigned
short which was 32 bits, pointers were 32 bits, but int was 128 bits. sizeof(storage_type) would be 64 bits, and the ints would not be aligned correctly. I think you need to define:
struct BB { struct B b; int i[1] };
typedef typename std::aigned_storage<sizeof(B),
alignof(B)>::type storage_type;

size_t length = 10;
char* storage = new char[sizeof(storage_type)+(length-1)*sizeof(int)];
B* pb = new(storage)B();
auto pbb = reinterpret_cast<BB*>(storage);
pb->p = new(&pbb->i)int[length];




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On Monday, December 14, 2015 at 8:00:20 AM UTC-5, Martin Bonner wrote:

On Sunday, 13 December 2015 19:10:14 UTC+1, Daniel wrote:

But what if we have

struct B
{
size_t length;
int *p;
};

and wish to allocate storage for B and p with placement new from the
same storage?

I think you need to define:
struct BB { struct B b; int i[1] };
typedef typename std::aigned_storage<sizeof(B),
alignof(B)>::type storage_type;

size_t length = 10;
char* storage = new char[sizeof(storage_type)+(length-1)*sizeof(int)];
B* pb = new(storage)B();
auto pbb = reinterpret_cast<BB*>(storage);
pb->p = new(&pbb->i)int[length];

Thank you very much! that definitely helps my understanding. But did you
mean to write BB instead of B in the typedef for storage_type?

Best regards,
Daniel


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On Tuesday, 15 December 2015 14:10:17 UTC+1, Daniel wrote:

On Monday, December 14, 2015 at 8:00:20 AM UTC-5, Martin Bonner wrote:

On Sunday, 13 December 2015 19:10:14 UTC+1, Daniel wrote:

But what if we have

struct B
{
size_t length;
int *p;
};

and wish to allocate storage for B and p with placement new from the
same storage?

I think you need to define:
struct BB { struct B b; int i[1] };
typedef typename std::aigned_storage<sizeof(B),
alignof(B)>::type storage_type;

size_t length = 10;
char* storage = new char[sizeof(storage_type) +
(length-1)*sizeof(int)];
B* pb = new(storage)B();
auto pbb = reinterpret_cast<BB*>(storage);
pb->p = new(&pbb->i)int[length];

Thank you very much! that definitely helps my understanding. But did you
mean to write BB instead of B in the typedef for storage_type?

Yes!

(and apologies to mod for failing to format correctly last time)


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On Sunday, December 13, 2015 at 12:10:14 PM UTC-6, Daniel wrote:

I'm trying to understand the rules for alignment with placement new.

If we have

struct A
{
size_t length;
};

my understanding (please correct me if I'm wrong) is that this results
in a
data structure that is properly aligned:

typedef typename std::aligned_storage<sizeof(A),
alignof(A)>::type storage_type;

char* storage = new char [sizeof(storage_type)];
A* pa = new(storage)A();

But what if we have

struct B
{
size_t length;
int *p;
};

and wish to allocate storage for B and p with placement new from the
same storage?

Wouldn't using std::aligned_union:

http://www.cplusplus.com/reference/type_traits/aligned_union/

be simpler? IOW:

using storage_type = aligned_union<Len,A,B>::type;

The only thing that needs to be calculated would be the Len
template parameter, which would just be the max of
the sizeof B and A.

HTH.

-regards,
Larry


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On Monday, January 11, 2016 at 6:50:17 AM UTC-6, evansl wrote:

On Sunday, December 13, 2015 at 12:10:14 PM UTC-6, Daniel wrote:

I'm trying to understand the rules for alignment with placement new.

If we have

struct A
{
size_t length;
};

my understanding (please correct me if I'm wrong) is that this results
in a
data structure that is properly aligned:

typedef typename std::aligned_storage<sizeof(A),
alignof(A)>::type storage_type;

char* storage = new char [sizeof(storage_type)];
A* pa = new(storage)A();

But what if we have

struct B
{
size_t length;
int *p;
};

and wish to allocate storage for B and p with placement new from the
same storage?

Wouldn't using std::aligned_union:

http://www.cplusplus.com/reference/type_traits/aligned_union/

be simpler? IOW:

using storage_type = aligned_union<Len,A,B>::type;

The only thing that needs to be calculated would be the Len
template parameter, which would just be the max of
the sizeof B and A.

[snip]
OOPS. The Len parameter to aligned_union must be a constant,
and it's not since it depends on length, which is a runtime
value.

Sorry for noise :(


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