Question

这是我的问题。我的kernel.h。

中有以下结构

struct   __Q_VECTOR__{
    double* Data;       
    int     Dimension;  
    int     Cluster;    
};

typedef struct __Q_VECTOR__     VQ_VECTOR;

在kernel.cu我有以下代码

int main(void){
 int L = 3, //.Data length
    N = 100;

VQ_VECTOR   *A,
            *device_VQ_VECTOR;

cudaError_t cudaStatus;

A =   (VQ_VECTOR*)malloc(N*sizeof(VQ_VECTOR));
for(int i=0; i<N; i++){
    VQ_VECTOR a;
    a.Data = (double*)malloc(L*sizeof(double));;
    a.Cluster   =   1;
    a.Dimension =   L;
    for(int j=0; j<L; j++)
        a.Data[j]=i*j;

    A[i] = a;
}

//Prinf of all the elements of A
for(int i=0; i<2; i++){
    printf("\nA[%d]={");
    for(int j=0; j<L; j++)
        printf("%.3f",A[i].Data[j]);
    printf("}");
}

printf("\n\n");
//I Allocate and Copy data from A to device_VQ_VECTORon the GPU memory

cudaDeviceReset();
cudaStatus = cudaMalloc((void**)&device_VQ_VECTOR, N*sizeof(VQ_VECTOR));
cudaStatus = cudaMemcpy(device_VQ_VECTOR, A, N*sizeof(VQ_VECTOR), cudaMemcpyHostToDevice);
cudaPrintfInit();
testKernel<<<N,1>>>(device_VQ_VECTOR, N);//to test and see on a sigle thread
cudaPrintfDisplay(stdout, true);
cudaPrintfEnd();
cudaStatus = cudaGetLastError();
if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "\n testKernel launch failed: %s\n", cudaGetErrorString(cudaStatus));
        return 1;
}
cudaStatus = cudaMemcpy(A, device_VQ_VECTOR, N*sizeof(VQ_VECTOR), cudaMemcpyDeviceToHost);
cudaStatus = cudaGetLastError();
if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "\n testKernel launch failed: %s\n", cudaGetErrorString(cudaStatus));
        return 1;
}
for(int i=0; i<2; i++){
    printf("\nA[%d]={");
    for(int j=0; j<L; j++)
        printf("%.3f",A[i].Data[j]);
    printf("}");
}
cudaFree(device_VQ_VECTOR);
 return 0;

}

当我建造时，有时它什么都不打印，有时它会起作用我的代码中有什么问题？可能是由

引起的

cudaStatus = cudaMalloc((void**)&device_VQ_VECTOR, N*sizeof(VQ_VECTOR));
cudaStatus = cudaMemcpy(device_VQ_VECTOR, A, N* sizeof(VQ_VECTOR), cudaMemcpyHostToDevice);

请帮忙！

Answer 1

这不起作用，因为数组是单独分配的，而不是复制到设备内存中。您还需要在设备上分配它们，并进行完整复制。更糟糕的是，您无法直接从主机端访问设备内存（除cudaMemcpy之外的其他方式），因此您无法使用例如cudaMalloc(&device_VQ_VECTOR[i].Data, ...)。 A[i].Data（它会崩溃）。

这是一个示例代码。为简单起见，它会丢弃主机端struct __Q_VECTOR__{ double* Data; int Dimension; int Cluster; }; typedef struct __Q_VECTOR__ VQ_VECTOR; __global__ void testKernel(VQ_VECTOR *X, int N){ int i= blockIdx.x*blockDim.x + threadIdx.x; cuPrintf("\n testKernel entrance by the global threadIdx= %d\n", i); for(int k=0; k<X[i].Dimension; k++) cuPrintf("%2.2f, ",X[i].Data[k]); cuPrintf("\n"); } int main(void){ int L = 3, //.Data length N = 100; VQ_VECTOR *A, *device_VQ_VECTOR; cudaError_t cudaStatus; A = (VQ_VECTOR*)malloc(N*sizeof(VQ_VECTOR)); for(int i=0; i<N; i++){ VQ_VECTOR a; a.Data = (double*)malloc(L*sizeof(double));; a.Cluster = 1; a.Dimension = L; for(int j=0; j<L; j++) a.Data[j]=(1+i)*(1+j); A[i] = a; } //Prinf of all the elements of A for(int i=0; i<2; i++){ printf("\nA[%d]={", i); for(int j=0; j<L; j++) printf("%.3f",A[i].Data[j]); printf("}\n"); } printf("\n\n"); //I Allocate and Copy data from A to device_VQ_VECTORon the GPU memory cudaDeviceReset(); cudaStatus = cudaMalloc((void**)&device_VQ_VECTOR, N*sizeof(VQ_VECTOR)); cudaStatus = cudaMemcpy(device_VQ_VECTOR, A, N*sizeof(VQ_VECTOR), cudaMemcpyHostToDevice); for(int i = 0; i != N; ++i) { /* can't access device_VQ_VECTOR[i].Data directly from host-side, * working around it with proxy variable */ double *out; cudaMalloc(&out, L*sizeof(double)); cudaMemcpy(out, A[i].Data, L*sizeof(double), cudaMemcpyHostToDevice); cudaMemcpy(&device_VQ_VECTOR[i].Data, &out, sizeof(void*), cudaMemcpyHostToDevice); // will re-allocate later, for simplicity sake free(A[i].Data); } cudaPrintfInit(); testKernel<<<N,1>>>(device_VQ_VECTOR, N);//to test and see on a sigle thread cudaPrintfDisplay(stdout, true); cudaPrintfEnd(); cudaStatus = cudaGetLastError(); if (cudaStatus != cudaSuccess) { fprintf(stderr, "\n testKernel launch failed: %s\n", cudaGetErrorString(cudaStatus)); return 1; } cudaStatus = cudaMemcpy(A, device_VQ_VECTOR, N*sizeof(VQ_VECTOR), cudaMemcpyDeviceToHost); for(int i = 0; i != N; ++i) { // allocate array, copy data double *array = (double*)malloc(L*sizeof(double)); cudaMemcpy(array, A[i].Data, L*sizeof(double), cudaMemcpyDeviceToHost); // assign new array to A[i] A[i].Data = array; } cudaStatus = cudaGetLastError(); if (cudaStatus != cudaSuccess) { fprintf(stderr, "\n testKernel launch failed: %s\n", cudaGetErrorString(cudaStatus)); return 1; } /* for(int i=0; i<2; i++){ printf("\nA[%d]={", i); for(int j=0; j<L; j++) printf("%.3f",A[i].Data[j]); printf("}\n"); }*/ cudaFree(device_VQ_VECTOR); // don't forget to free A and all its Data return 0; }，然后重新创建它们。这不太好，但那会好转。

[2, 0]: 3.00, [18, 0]: 19.00, [22, 0]: 23.00, [16, 0]: 17.00,
[24, 0]: 25.00, [19, 0]: 20.00, [4, 0]: 5.00, [23, 0]: 24.00,
[3, 0]: 4.00, [5, 0]: 6.00, [13, 0]: 14.00, [1, 0]: 2.00,
[10, 0]: 11.00, [6, 0]: 7.00, [14, 0]: 15.00, [0, 0]: 1.00, [20, 0]:

部分输出将是（它很大，我不想发布太多）：

{{1}}

cuda将用户定义的结构传递给内核失败

1 个答案: