Question

我的CUDA内核需要很多需要作为内核指针传递的数组。问题是在内核启动之前，所有指针都有有效的地址，而且cudaMalloc和cudaMemcpy调用总是返回cudaSuccess，但是一旦内核启动，所有这些参数都变为null ！

我对发生的事情毫无头绪。这是我使用cuda-gdb

运行代码时得到的结果

CUDA Exception: Device Illegal Address
The exception was triggered in device 0.

Program received signal CUDA_EXCEPTION_10, Device Illegal Address.
[Switching focus to CUDA kernel 0, grid 1, block (0,0,0), thread (64,0,0), device 0, sm 1, warp 2, lane 0]
0x00000000062a3dd8 in compute_data_and_match_kernel<<<(2,1,1),(512,1,1)>>> (a11=0x0, a12=0x0, a22=0x0, b1=0x0, b2=0x0, mask=0x0, wx=0x0, wy=0x0, du=0x0, dv=0x0, uu=0x0, 
    vv=0x0, Ix_c1=0x0, Ix_c2=0x0, Ix_c3=0x0, Iy_c1=0x0, Iy_c2=0x0, Iy_c3=0x0, Iz_c1=0x0, Iz_c2=0x0, Iz_c3=0x0, Ixx_c1=0x0, Ixx_c2=0x0, Ixx_c3=0x0, Ixy_c1=0x0, Ixy_c2=0x0, 
    Ixy_c3=0x0, Iyy_c1=0x0, Iyy_c2=0x0, Iyy_c3=0x0, Ixz_c1=0x0, Ixz_c2=0x0, Ixz_c3=0x0, Iyz_c1=0x0, Iyz_c2=0x0, Iyz_c3=0x0, desc_weight=0x0, desc_flow_x=0x0, 
    desc_flow_y=0x0, half_delta_over3=0.0833333358, half_beta=0, half_gamma_over3=0.833333313, width=59, height=26, stride=60) at opticalflow_aux.cu:441
441         ix_c1_val = Ix_c1[index]; iy_c1_val = Iy_c1[index]; iz_c1_val = Iz_c1[index];
(cuda-gdb)

是否有一些我非常遗憾的事情。提前谢谢。

编辑1： 正如Gilles所建议的那样，我试图将主机指针和数据复制到结构中，然后复制到设备上。为简单起见（MCVE）我在struct中只使用了一个指针：

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

typedef struct test {
    float *ptr;
} test_t;

__global__ void test_kernel(test_t *s) {
    s->ptr[0] = s->ptr[1] = s->ptr[2] = s->ptr[3] = s->ptr[4] = 100;
    s->ptr[5] = s->ptr[6] = s->ptr[7] = s->ptr[8] = s->ptr[9] = 100;
}

int main() {

    float arr[] = {0,1,2,3,4,5,6,7,8,9};

    test_t *h_struct;
    h_struct = (test_t *)malloc(sizeof(test_t));
    h_struct->ptr = arr;

    test_t *d_struct;
    float *d_data;
    cudaMalloc((void **)&d_struct, sizeof(test_t));
    cudaMalloc((void **)&d_data, sizeof(float)*10);

    // Copy the data from host to device
    cudaMemcpy(d_data, h_struct->ptr, sizeof(float)*10,   cudaMemcpyHostToDevice);
    // Point the host struct ptr to device memory
    h_struct->ptr = d_data;
    // copy the host struct to device
    cudaMemcpy(d_struct, h_struct, sizeof(test_t), cudaMemcpyHostToDevice);


    // Kernel Launch
    test_kernel<<<1,1>>>(d_struct);
    // copy the device array to host
    cudaMemcpy(h_struct->ptr, d_data, sizeof(float)*10, cudaMemcpyDeviceToHost);

    cudaFree(d_data);
    cudaFree(d_struct);

    // Verifying if all the values have been set to 100
    int i;
    for(i=0 ; i<10 ; i++)
        printf("%f\t", h_struct->ptr[i]);

    return 0;
}

当我检查d_struct->ptr的值时，就在内核启动之前，它会显示0x0。（我在调试模式下使用nsight检查了这些值）

Answer 1

不确定是否是问题，但我相信将参数传递给内核的堆栈大小是有限的。您可能需要创建一个存储参数的结构，将其复制到设备并仅将指针作为参数传递给内核。然后，在内核中从结构中检索你的参数......

编辑：添加了已提交代码的更正版本。这对我有用，并举例说明了我所描述的原则。

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

typedef struct test {
    float *ptr;
} test_t;

__global__ void test_kernel(test_t *s) {
    s->ptr[0] = s->ptr[1] = s->ptr[2] = s->ptr[3] = s->ptr[4] = 100;
    s->ptr[5] = s->ptr[6] = s->ptr[7] = s->ptr[8] = s->ptr[9] = 100;
}

int main() {

    float arr[] = {0,1,2,3,4,5,6,7,8,9};

    test_t *h_struct;
    h_struct = (test_t *)malloc(sizeof(test_t));

    test_t *d_struct;
    float *d_data;
    cudaMalloc((void **)&d_struct, sizeof(test_t));
    cudaMalloc((void **)&d_data, sizeof(float)*10);

    // Copy the data from host to device
    cudaMemcpy(d_data, arr, sizeof(float)*10, cudaMemcpyHostToDevice);
    // Point the host struct ptr to device memory
    h_struct->ptr = d_data;
    // copy the host struct to device
    cudaMemcpy(d_struct, h_struct, sizeof(test_t), cudaMemcpyHostToDevice);

    // Kernel Launch
    test_kernel<<<1,1>>>(d_struct);
    // copy the device array to host
    cudaMemcpy(arr, d_data, sizeof(float)*10, cudaMemcpyDeviceToHost);

    cudaFree(d_data);
    cudaFree(d_struct);

    // Verifying if all the values have been set to 100
    int i;
    for(i=0 ; i<10 ; i++)
        printf("%f\t", arr[i]);

    return 0;
}

CUDA内核指针参数变为NULL

1 个答案: