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CUDA allocation alignment is 256 bytes - seriously?

Tags:

cuda

textures

In "CUDA C Programming Guide 5.0", p73 (also here) says "Any address of a variable residing in global memory or returned by one of the memory allocation routines from the driver or runtime API is always aligned to at least 256 bytes". I do not know the exact meaning of this sentence. Could anyone show an example for me? Many thanks.

A derivative question: So, what about allocating an one-dimensional array of basic elements (like int) or self-defined ones? The starting address of the array will be multiples of 256B, while the address of each element in the array is not necessarily multiples of 256B?

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jsc0218 Avatar asked Dec 29 '12 15:12

jsc0218


1 Answers

The pointers which are allocated by using any of the CUDA Runtime's device memory allocation functions e.g cudaMalloc or cudaMallocPitch are guaranteed to be 256 byte aligned, i.e. the address is a multiple of 256.

Consider the following example:

char *ptr1, *ptr2;

int bytes = 1;

cudaMalloc((void**)&ptr1,bytes);
cudaMalloc((void**)&ptr2,bytes);

Suppose the address returned in ptr1 is some multiple of 256, then the address returned in ptr2 will be atleast (ptr1 + 256).

This is a restriction imposed by the device on which the memory is being allocated. Mostly, pointers are aligned due to performance purposes. (Some NVIDIA guy should be able to tell if there is some other reason also).

Important:

Pointer alignment is not always 256. On my device (GTX460M), it is 512. You can get the device pointer alignment by the cudaDeviceProp::textureAlignment field.

Alignment of pointers is also a requirement for binding the pointer to textures.

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sgarizvi Avatar answered Nov 17 '22 02:11

sgarizvi