CODEWITHSUNDEEP
visual-c++memory-alignment
ledokol
ledokol

Reputation: 133

Types default alignment in visual C++

Just noticed something that looks strange to me. Visual C++ doesn't align object in their required boundary by default. For example long long is aligned to 4 bytes boundary, while __alignof(T) returns 8 (as far as I see it always returns size of the type). So it looks like it is not properly aligned. For example

long long a1;
char g;
long long a2;
// alignment check for &a2 fails
if (((uintptr_t)&a2 & (__alignof(long long) - 1)) != 0) // failed

I also tried just to print the pointer, the value of &a2 is 0x0035F8FC(3537148 in dec).

Is there something I get wrong? I need properly aligned object of type long long. What can I do about that? I could use __declspec(align()) Microsoft extension, but it requires literal number, so I can't write anything like that.

__declspec(align(__alignof(long long))) long long object;

Upvotes: 1

Views: 2496

Answers (2)

Necrolis
Necrolis

Reputation: 26171

VC doesn't guarantee automatic stack alignment of variables, at most the variable will be aligned to the stacks alignment(generally 4 bytes on 32 bit systems). If you need special alignment, you need to use __declspec(align(x)), just like MSVC's SSE types(like __m128), else you'll need to use _aligned_malloc instead

Upvotes: 4

harper
harper

Reputation: 13690

The alignment should minimize the memory cycles on RAM access. The 4 byte alignment uses only two 32-bit acccesses to the long long. The 8 byte aligment doesn't improve the behavior. The compiler has a default alignment that can be overwritten with the /Zp option.

See also: Configuration Properties C/C++ Code Generation Struct Member Aligment.

Upvotes: 0

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