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为什么 cuMemAddressReserve() 因 CUDA_INVALID_VALUE 而失败?

[英]Why is cuMemAddressReserve() failing with CUDA_INVALID_VALUE?

Consider the following program (written in C syntax):考虑以下程序(用 C 语法编写):

#include <cuda.h>
#include <stdio.h>
#include <stdlib.h>

int main() {
    CUresult result;
    unsigned int init_flags = 0;
    result = cuInit(init_flags);
    if (result != CUDA_SUCCESS) { exit(EXIT_FAILURE); }
    CUcontext ctx;
    unsigned int ctx_create_flags = 0;
    CUdevice device_id = 0;
    result = cuCtxCreate(&ctx, ctx_create_flags, device_id);
    // Note: The created context is also made the current context,
    // so we are _in_ a context from now on.
    if (result != CUDA_SUCCESS) { exit(EXIT_FAILURE); }
    CUdeviceptr requested = 0;
    CUdeviceptr reserved;
    size_t size = 0x20000;
    size_t alignment = 0; // default
    unsigned long long reserve_flags = 0;

    // -----------------------------------
    // ==>> FAILURE on next statement <<==
    // -----------------------------------

    result = cuMemAddressReserve(&reserved, size, alignment, requested, reserve_flags);
    if (result != CUDA_SUCCESS) {
        const char* error_string;
        cuGetErrorString(result, &error_string);
        fprintf(stderr, "cuMemAddressReserve() failed: %s\n", error_string);
        exit(EXIT_FAILURE);
    }
    return 0;
}

This fails when trying to make the reservation:尝试预订时失败:

cuMemAddressReserve() failed: invalid argument

what's wrong with my arguments?我的 arguments 怎么了? Is it the size?是尺码吗? the alignment? alignment? Requesting an address of 0?请求地址为 0? If it's the latter - how can I even know what address to request, when I don't really care?如果是后者 - 当我真的不在乎的时候,我怎么知道要请求哪个地址?

If I recall correctly,the sizes for virtual memory management functions must be a multiple of CUDAs allocation granularity.如果我没记错的话,虚拟 memory 管理函数的大小必须是 CUDA 分配粒度的倍数。 See cuMemGetAllocationGranularity and this blog post https://developer.nvidia.com/blog/introducing-low-level-gpu-virtual-memory-management/请参阅cuMemGetAllocationGranularity和这篇博文 https://developer.nvidia.com/blog/introducing-low-level-gpu-virtual-memory-management/

The following works on my machine.以下适用于我的机器。

#include <cuda.h>
#include <stdio.h>
#include <stdlib.h>

int main() {
    CUresult result;
    unsigned int init_flags = 0;
    result = cuInit(init_flags);
    if (result != CUDA_SUCCESS) { exit(EXIT_FAILURE); }
    CUcontext ctx;
    unsigned int ctx_create_flags = 0;
    CUdevice device_id = 0;
    result = cuCtxCreate(&ctx, ctx_create_flags, device_id);
    // Note: The created context is also made the current context,
    // so we are _in_ a context from now on.
    if (result != CUDA_SUCCESS) { exit(EXIT_FAILURE); }
    CUdeviceptr requested = 0;
    CUdeviceptr reserved;
    size_t size = 0x20000;
    size_t alignment = 0; // default
    unsigned long long reserve_flags = 0;

    size_t granularity;
    CUmemAllocationProp prop;
    prop.type = CU_MEM_ALLOCATION_TYPE_PINNED;
    prop.location.type = CU_MEM_LOCATION_TYPE_DEVICE;
    prop.location.id = (int)0;
    prop.win32HandleMetaData = NULL;
    result = cuMemGetAllocationGranularity (&granularity, &prop, CU_MEM_ALLOC_GRANULARITY_MINIMUM );
    if (result != CUDA_SUCCESS) { exit(EXIT_FAILURE); }
    printf("minimum granularity %lu\n", granularity);


    size_t padded_size = ((granularity + size - 1) / granularity) * granularity;
    result = cuMemAddressReserve(&reserved, padded_size, alignment, requested, reserve_flags);
    if (result != CUDA_SUCCESS) {
        const char* error_string;
        cuGetErrorString(result, &error_string);
        fprintf(stderr, "cuMemAddressReserve() failed: %s\n", error_string);
        exit(EXIT_FAILURE);
    }
    return 0;
}

tl;dr: Your reserved region size is not a multiple of (some device's) allocation granularity. tl; dr:您的保留区域大小不是(某些设备的)分配粒度的倍数。

As @AbatorAbetor suggested, cuMemAddressReserve() implicitly requires the size of the memory region to be a multiple of some granularity value.正如@AbatorAbetor 建议的那样, cuMemAddressReserve()隐式要求 memory 区域的大小是某个粒度值的倍数。 And despite 0x20000 seeming like a generous enough value for that (2^21 bytes... system memory pages are typically 4 KiB = 2^12 bytes) - NVIDIA GPUs are very demanding here.尽管 0x20000 似乎是一个足够大的值(2^21 字节......系统 memory 页面通常为 4 KiB = 2^12 字节) - NVIDIA GPU 在这里非常苛刻。

For example, a Pascal GTX 1050 Ti GPU with ~4GB of memory has a granularity of 0x200000, or 2 MiB - 16 times more than what you were trying to allocate.例如,Pascal GTX 1050 Ti GPU 和 ~4GB memory 的粒度为 0x200000,或 2 MiB - 比您尝试分配的多 16 倍。

Now, what would happen if we had two devices with different granularity values?现在,如果我们有两个具有不同粒度值的设备会发生什么? Would we need to use the least-common-multiple?我们需要使用最小公倍数吗? Who knows.谁知道。

Anyway, bottom line: Always check the granularity both before allocating and before reserving.无论如何,底线:始终在分配之前和保留之前检查粒度。

I have filed this as a documentation bug with NVIDIA, bug 3486420 (but you may not be able to follow the link, because NVIDIA hide their bugs from their users).我已将此作为文档错误提交给 NVIDIA,错误 3486420 (但您可能无法访问该链接,因为 NVIDIA 对其用户隐藏了他们的错误)。

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