cuBlas、cuda 功能不起作用，没有产生错误

Question

这是一个说明问题的最小程序。我使用的是具有 128 个 CUDA 核心的 GTS 250，Windows 7 上的 CUDA 5.0。

void cuda_ops_test(int N, float* R)
{
    //Values of input matrix in CPU
    fprintf(stderr, "\nValues of R: \n");
    for (int i=0; i<N; ++i)
        fprintf(stderr, "%f, ", R[i]);
    fprintf(stderr, "\n");

    //Initialize CUDA/CUBLAS
    cublasHandle_t handle;
    cublasStatus_t status;
    status = cublasCreate(&handle);
    if (status == CUBLAS_STATUS_SUCCESS)
        fprintf(stderr, "CUBLAS initialization succeeded.\n");

    //Allocate device memory
    float *dR = 0;
    cudaError_t alloc_status;
    alloc_status = cudaMalloc((void**)&dR, N*sizeof(dR[0]));
    if(alloc_status == cudaSuccess)
        fprintf (stderr, "\nDevice memory allocation succeeded.\n");

    //Load array into memory
    status = cublasSetMatrix(1, N, sizeof(R[0]), R, N, dR, N);
    if(status == CUBLAS_STATUS_SUCCESS)
        fprintf (stderr, "\nDevice write succeeded.\n");

    //First operation: summation
    float ans;
    status = cublasSasum(handle, N, dR, 1, &ans);
    if (status == CUBLAS_STATUS_SUCCESS)
        fprintf(stderr, "\ncublasSasum produced no error. Sum of dR: %d\n", ans);
    else
        fprintf(stderr, "\ncublasSasum error: %d.\n", status);

    //Second operation: y = ax+y
    const float alpha = 2.0;
    status = cublasSaxpy(handle, N,
                 &alpha,
                 dR, 1,
                 dR, 1);
    if (status == CUBLAS_STATUS_SUCCESS)
        fprintf(stderr, "\ncublasSaxpy produced no error.\n");
    else
        fprintf(stderr, "\ncublasSaxpy error: %d.\n", status);

    // transfer device dR to host R
    status = cublasGetMatrix (1, N, sizeof(dR[0]), dR, N, R, N);
    if(status == CUBLAS_STATUS_SUCCESS)
        fprintf (stderr, "\nDevice read succeded\n");

    //Display post-op values of R
    fprintf(stderr, "\nValues of R, after cublasSaxpy: \n");
    for (int i=0; i<N; ++i)
        fprintf(stderr, "%f, ", R[i]);
    fprintf(stderr, "\n");

    //Attempt to zero with cudaMemset
    cudaError_t stat = cudaMemset(dR, 0, N*sizeof(dR[0]));
    if (stat==cudaSuccess)
        fprintf(stderr, "\nZeroing with cudaMemset on R produced no error.\n");

    //Again transfer device dR to host R, after zeroing
    status = cublasGetMatrix (1, N, sizeof(dR[0]), dR, N, R, N);
    if(status == CUBLAS_STATUS_SUCCESS)
        fprintf (stderr, "\nDevice read succeded.\n");

    //Display values of R again
    fprintf(stderr, "\nValues of R, after zeroing with cudaMemset: \n");
    for (int i=0; i<N; ++i)
        fprintf(stderr, "%f, ", R[i]);
    fprintf(stderr, "\n");

    cudaFree(dR);
}

以下是输出，表明当数据加载到 GPU 内存中时，实际上没有发生任何操作:

Values of R: 0.123020, 0.367809, 0.834681, 0.035096, 0.517014, 0.662984, 0.426221, 0.104678,

CUBLAS initialization succeeded.

Device memory allocation succeeded.

cublasSasum produced no error. Sum of dR: 0

cublasSaxpy produced no error.

Values of R, after cublasSaxpy: 0.123020, 0.367809, 0.834681, 0.035096, 0.517014, 0.662984, 0.426221, 0.104678,

Zeroing with cudaMemset on R produced no error.

Values of R, after zeroing with cudaMemset: 0.123020, 0.367809, 0.834681, 0.035096, 0.517014, 0.662984, 0.426221, 0.104678,

发生什么事了？ (节日快乐。:) )

Answer 1

您的代码有几个错误。正如我在评论中提到的，您错过了这样一个事实:您的设备读取和设备写入消息都没有被打印出来，因为这些函数(cublasSetMatrix、cublasGetMatrix)实际上失败了。

要修复 cublasSetMatrix 和 cublasGetMatrix 调用，请将 lda 和 ldb 参数更改为 1:

 status = cublasSetMatrix(1, N, sizeof(R[0]), R, 1, dR, 1);    
 ...     
 status = cublasGetMatrix (1, N, sizeof(dR[0]), dR, 1, R, 1);

documentation对于这些函数来说:“源矩阵 A 和目标矩阵 B 的前导维度分别在 lda 和 ldb 中给出。前导维度表示分配的矩阵的行数”

在打印 cublasSasum 操作结果的行中，您的 printf 语句错误地使用 int 格式说明符来打印浮点值。这行不通。将 %d 更改为 %f :

 fprintf(stderr, "\ncublasSasum produced no error. Sum of dR: %f\n", ans);

通过这些更改，我能够得到合理的结果:

Values of R:
0.123020, 0.367809, 0.834681, 0.035096, 0.517014, 0.662984, 0.426221, 0.104678,
CUBLAS initialization succeeded.

Device memory allocation succeeded.

Device write succeeded.

cublasSasum produced no error. Sum of dR: 3.071503

cublasSaxpy produced no error.

Device read succeded

Values of R, after cublasSaxpy:
0.369060, 1.103427, 2.504043, 0.105288, 1.551042, 1.988952, 1.278663, 0.314034,

Zeroing with cudaMemset on R produced no error.

Device read succeded.

Values of R, after zeroing with cudaMemset:
0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000,

请注意 this SO question/answer提供了有关有用、方便的 cublas 错误解析器函数的提示。将其构建到 cublas 函数调用的包装器或错误检查宏中并不困难。