Re: [eigen] STL Vectors and Alignment

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On 26.11.2014 17:39, Gabriel wrote:

Thanks for the answer!

Sorry again for my unclear phrasing!
Just to be absolutely sure:

Lets say we have:

     struct A{
          Eigen::Vector2d t;
     };

If we use A locally:
     A a;
     a.t
is a.t  aligned on the Stack, (how does alignment work on the stack)

Yes, that is aligned on the stack. Also, if you wrote, e.g.
  struct A{
    char dummy;      // compiler will add filling bytes after this
    Eigen::Vector2d t;
  };
a.t will be aligned on the stack.
The data member of Vector2d has a compiler attribute which tells it to always have it aligned.

If we use A on the heap:
     A * a = new A;
then as "a" is not aligned , also a.t is not aligned

It is not guaranteed to be aligned. It is, however, aligned on most (all?) 64bit systems.

If we use EIGEN_MAKE_ALLOCATOR_NEW inside A , then "a" is aligned , and

Yes.

BUT what happens in this example:

     struct B{
          A a;
     };
     struct C{
          B b;
     };

     std::vector<   C  ,   Eigen::aligned_allocator<C>  > vec(1);

Is      vec[0].b.a.t      now  aligned in memory? which does *NOT*
result in a segfault if Eigen uses SIMD internally?

Yes.

If we use locally lets say:

     C c;
     c.b.a.t  do something with t

is c.b.a.t  aligned on the Stack?

Yes.

Or on the heap:

     C * c = new C;

c might not be aligned => c.b  might not be aligned  => c.b.a => might
not bealigned  => c.b.a.t might not be aligned  => SEGFAULT for usage of
c.b.a.t

Correct. c is not guaranteed to be aligned (except on most 64 bit systems ...)

How do I need to correct the above example such that in the cases above,
vector A::t is always aligned to 16bytes?

To make the only not yet working case work, you need to add an
  EIGEN_MAKE_ALIGNED_OPERATOR_NEW
into struct C. It would not hurt to add it into A and B, as well.


Christoph

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Dipl.-Inf., Dipl.-Math. Christoph Hertzberg
Cartesium 0.049
Universität Bremen
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