CUDA - 将设备数据复制到主机? [英] CUDA - Copy device data to host?

问题描述

我有设备变量，在这个变量中，我在设备中分配并填充了一个数组，但是我在将数据获取到主机时遇到了问题.cudaMemcpy() 返回 cudaErrorInvalidValue 错误.我该怎么做?

I have device variable and in this variable, I allocate and fill an array in the device, but I have a problem to get data to host. cudaMemcpy() return cudaErrorInvalidValue error. how can I do it?

PS:代码只是示例，我知道，在这种特殊情况下，我可以使用 cudaMalloc 因为我知道数组的大小，但在我的真实代码中，它计算的大小设备中的数组，它需要立即分配内存.

PS: The Code is just example, I know, that In this particular case I can use cudaMalloc because I know the size of the array, but In my REAL code, It computes the size of the array in the device and it needs immediately allocate memory.

PS2:我发现了一个类似的问题，但我还是不知道，我该如何解决呢?- 从设备复制在设备中分配的数据主持

PS2: I found a similar problem, but I still don't know, how can I solve it? - copy data which is allocated in device from device to host

PS3:我更新了代码，但还是不行:{

PS3: I have updated code, but still doesn't work:{

PS4:我只是想在装有 Nvidia GT 520MX(最新游戏驱动程序)的笔记本上运行此代码，但也无法正常工作:(

PS4: I am just trying to run this code on a notebook with Nvidia GT 520MX(latest game driver) and doesn't work too :(

谢谢

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

#define N 400
__device__ int* d_array;

__global__ void allocDeviceMemory()
{
    d_array = new int[N];
    for(int i=0; i < N; i++)
         d_array[i] = 123;
}

int main()
{
    allocDeviceMemory<<<1, 1>>>();

    cudaDeviceSynchronize();

    int* d_a = NULL;
    cudaMemcpyFromSymbol((void**)&d_a, "d_array", sizeof(d_a), 0, cudaMemcpyDeviceToHost);
    printf("gpu adress: %lld
", d_a);


    int* h_array = (int*)malloc(N*sizeof(int));
    cudaError_t errr = cudaMemcpy(h_array, d_a, N*sizeof(int), cudaMemcpyDeviceToHost);
    printf("h_array: %d, %d
", h_array[0], errr);

    getchar();
    return 0;
}

推荐答案

我已经测试了你的代码，这里没有错误.我正在运行 CUDA 4.0.

i have tested your code and there is no error here. I am running CUDA 4.0.

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