Re: Make STL containers allocate aligned memory

From:
Maxim Yegorushkin <maxim.yegorushkin@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Thu, 27 Nov 2008 03:51:34 -0800 (PST)
Message-ID:
<78900f33-df74-4850-8fd3-d795b8ed7457@j35g2000yqh.googlegroups.com>
On Nov 26, 6:06 pm, zr <zvir...@gmail.com> wrote:

On Nov 26, 7:32 pm, Maxim Yegorushkin <maxim.yegorush...@gmail.com>
wrote:

On Nov 26, 5:03 pm, Maxim Yegorushkin <maxim.yegorush...@gmail.com>
wrote:

On Nov 26, 4:42 pm, zr <zvir...@gmail.com> wrote:

I need to use STL containers that allocate aligned memory.


Why?

My compiler provides an aligned_malloc routine.


It is probably provided by a library shipped with the compiler, not b=

y

compiler itself.

How can this be accomplished?


Something like this:

#include <memory>
#include <stdexcept>

template<class T, size_t alignment>
struct aligned_allocator : std::allocator<T>
{
    template<class U>
    struct rebind { typedef aligned_allocator<U, alignment> other=

; };

    typedef std::allocator<T> base;

    typedef typename base::pointer pointer;
    typedef typename base::size_type size_type;

    pointer allocate(size_type n)
    {
        if(pointer p = (pointer)aligned_malloc(n, alignment=

))

            return p;
        throw std::bad_alloc("aligned_allocator");
    }

    pointer allocate(size_type n, void const*)
    {
        return this->allocate(n);
    }

    void deallocate(pointer p, size_type)
    {
        aligned_free(p);
    }

};


Usage is:

    typedef std::vector<X, aligned_allocator<X, required_alignment>=

 >

VecX;

--
Max


Max, thanks. I wrote a quick test on your proposal. Unfortunately, got
a compilation error:
C:\Program Files (x86)\Microsoft Visual Studio 9.0\VC\include\xutility
(419) : error C2664: 'aligned_allocator<T,alignment>::aligned_allocator
(const aligned_allocator<T,alignment> &) throw()' : cannot convert
parameter 1 from 'aligned_allocator<T,alignment>' to 'const
aligned_allocator<T,alignment> &'
        with
        [
            T=std::_Aux_cont,
            alignment=64
        ]
        and
        [
            T=int,
            alignment=64
        ]
        and
        [
            T=std::_Aux_cont,
            alignment=64
        ]
        Reason: cannot convert from 'aligned_allocator<T,alignmen=

t>'

to 'const aligned_allocator<T,alignment>'
        with
        [
            T=int,
            alignment=64
        ]
        and
        [
            T=std::_Aux_cont,
            alignment=64
        ]
        No user-defined-conversion operator available that can pe=

rform

this conversion, or the operator cannot be called
        C:\Program Files (x86)\Microsoft Visual Studio 9.0\VC\inc=

lude

\xutility(417) : while compiling class template member function
'std::_Container_base_aux_alloc_real<_Alloc>::_Container_base_aux_alloc_r=

eal

(_Alloc)'
        with
        [
            _Alloc=aligned_allocator<int,64>
        ]
        C:\Program Files (x86)\Microsoft Visual Studio 9.0\VC\inc=

lude

\vector(421) : see reference to class template instantiation
'std::_Container_base_aux_alloc_real<_Alloc>' being compiled
        with
        [
            _Alloc=aligned_allocator<int,64>
        ]
        C:\Program Files (x86)\Microsoft Visual Studio 9.0\VC\inc=

lude

\vector(439) : see reference to class template instantiation
'std::_Vector_val<_Ty,_Alloc>' being compiled
        with
        [
            _Ty=int,
            _Alloc=aligned_allocator<int,64>
        ]
        .\main.cpp(38) : see reference to class template instanti=

ation

'std::vector<_Ty,_Ax>' being compiled
        with
        [
            _Ty=int,
            _Ax=aligned_allocator<int,64>
        ]

Following is the source of the test:

#include <malloc.h>
#include <memory>
#include <stdexcept>
#include <vector>

template<class T, size_t alignment>
struct aligned_allocator : std::allocator<T>
{
    template<class U>
    struct rebind { typedef aligned_allocator<U, alignment> other; };

    typedef std::allocator<T> base;

    typedef typename base::pointer pointer;
    typedef typename base::size_type size_type;

    pointer allocate(size_type n)
    {
        if(pointer p = (pointer)_aligned_malloc(n, alignment))
            return p;
        throw std::bad_alloc("aligned_allocator");
    }

    pointer allocate(size_type n, void const*)
    {
        return this->allocate(n);
    }

    void deallocate(pointer p, size_type)
    {
        _aligned_free(p);
    }

};

int main(int argc, char* argv[])
{
        std::vector<int, aligned_allocator<int, 64> > v(1024);
}


Strange, it compiles fine with M$VC 2005. It looks like
aligned_allocator needs a conversion constructor. Try adding these
constructors:

    aligned_allocator()
    {}

    template<class U>
    aligned_allocator(U const&)
    {}

--
Max

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