__m128i requires 16-byte alignment, which vector doesn't guarantee to
provide (well, it should do, but __m128i has especially strict alignment
requirements - ones that aren't met by malloc). One way of working
around this would be to specify a special vector allocator which did
align the memory correctly. e.g. (VC2005 specific due to __alignof())

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

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

    aligned_allocator(){}

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

    template <class U>
    aligned_allocator& operator=(aligned_allocator<U> const& other){
        return *this;
    }

    T* allocate(std::size_t N, const void* hint)
    {
        return allocate(N);
    }

    T* allocate(size_type N)
    {
        return static_cast<T*>(_aligned_malloc(N * sizeof(T), __alignof(T)));
    }

    void deallocate(T* p, std::size_t N)
    {
        _aligned_free(p);
    }
};

Then:
std::vector<__m128i, aligned_allocator<__m128i> > list;

Tom