Dynamic allocated memory not aligned in SSE

Question

Here's code which works normally:

char a[100];
for (int i = 0; i < 100; i++)
     a[i] = 0;
__m128i x = _mm_load_si128((__m128i *) a);

But if I dynamically allocate memory, VS 2013 will interrupt:

char *a = new char[100];
for (int i = 0; i < 100; i++)
     a[i] = 0;
__m128i x = _mm_load_si128((__m128i *) a);

How can I use both dynamic memory and aligned load instruction?

I think I totally misunderstood the question. Is there an error message or is the memory actually not aligned as desired? — Codor, Jan 13 '15 at 14:45
The second one interrupt if I use the aligned load instruction. — KUN, Jan 13 '15 at 14:48
I checked your code and it work fine on my console application. No errors. — LPs, Jan 13 '15 at 14:59

score 2 · Answer 1 · answered Jan 13 '15 at 14:44

2

Beacuse of sizeof is returning the size of a char pointer: 4 bytes.

answered Jan 13 '15 at 14:44

LPs

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Codor · Answer 2 · 2015-01-13T15:01:02.643

You could either manipulate the new operator to return aligned memory (which might not be the best idea maintainance-wise), or you might allocate more memory and just operate on the aligned part of it. If, for instance, you need an alignment on an 8-byte even address, allocate 8 bytes more and work on an aligned pointer as follows.

char* a;
uintptr_t unaligned = (uintptr_t)a;
uintptr_t aligned = ( unaligned % 8 ) ? unaligned : ( unaligned % 8 ) + 8;
char* a_aligned = (char*)aligned;

The type uintptr_t is defined in stdint.h.

Dynamic allocated memory not aligned in SSE

2 Answers2