c++ - CUBLAS 矩阵乘法与行主数据无转置

Question

我目前正尝试在我的 GPU 上使用 CUBLAS 实现矩阵乘法。

它适用于方矩阵和特定大小的输入，但对于其他输入，最后一行不会返回(并且包含 0，因为这是我实现它的方式)。

我认为这是 cublasSgemm 的分配或语法问题，但我找不到它在哪里。

注意:如果您不熟悉 CUBLAS:它是 column-majored ，这就是为什么看起来操作是以另一种方式执行的。

如有任何帮助，我们将不胜感激。

---

代码:

请注意，gpuErrchk 和 cublasErrchk 当然与此处无关。

#include <cuda.h>
#include <cuda_runtime.h>
#include <cublas_v2.h>

#include <vector>

std::vector<float> CUDA_mult_MAT(const std::vector<float> &data_1 , const uint64_t data_1_rows, const uint64_t data_1_columns,
                                 const std::vector<float> &data_2 , const uint64_t data_2_rows, const uint64_t data_2_columns){

    cublasHandle_t handle;

    cublasErrchk(cublasCreate(&handle));

    std::vector<float> result(data_1_rows * data_2_columns); //Vector holding the result of the multiplication

    /*----------------------------------------------------------------------------------------------*/

    float* GPU_data_1 = NULL;
    gpuErrchk(cudaMalloc((void**)&GPU_data_1 , data_1.size()*sizeof(float))); //Allocate memory on the GPU
    gpuErrchk(cudaMemcpy(GPU_data_1, data_1.data(), data_1.size()*sizeof(float), cudaMemcpyHostToDevice)); //Copy data from data_1 to GPU_data_1

    float* GPU_data_2 = NULL;
    gpuErrchk(cudaMalloc((void**)&GPU_data_2 ,data_2.size()*sizeof(float))); //Allocate memory on the GPU
    gpuErrchk(cudaMemcpy(GPU_data_2, data_2.data(), data_2.size()*sizeof(float), cudaMemcpyHostToDevice));//Copy data from data_2 to GPU_data_2

    float* GPU_result = NULL;
    gpuErrchk(cudaMalloc((void**)&GPU_result , result.size()*sizeof(float))); //Allocate memory on the GPU

    /*----------------------------------------------------------------------------------------------*/


    const float alpha = 1.f; 
    const float beta = 0.f;

    cublasErrchk(
               cublasSgemm(handle , CUBLAS_OP_N , CUBLAS_OP_N,
                           data_2_columns , data_2_rows ,data_1_columns,
                           &alpha , GPU_data_2 , data_2_columns,
                           GPU_data_1 , data_1_columns,
                           &beta , GPU_result , data_1_rows)
             ); //Perform multiplication 



    gpuErrchk(cudaMemcpy(result.data() , GPU_result , result.size() * sizeof(float) , cudaMemcpyDeviceToHost)); //Copy back to the vector 'result'

    gpuErrchk(cudaFree(GPU_data_1)); //Free GPU memory
    gpuErrchk(cudaFree(GPU_data_2)); //Free GPU memory
    gpuErrchk(cudaFree(GPU_result)); //Free GPU memory

    cublasErrchk(cublasDestroy_v2(handle)); 


    return result;


}

---

输入:

#include <iostream>

#include <vector>

int main(){

    const std::vector<float> r1 =  CUDA_mult_MAT({1 , 2 , 3 , 4 , 5 , 6} , 2 , 3 ,
                                           {7 , 8 , 9 , 10 , 11 , 12} , 3 , 2);
/*
Product :
         7  8
1 2 3    9  10
4 5 6    11 12

*/

    for(auto & value: r1){std::cout << value << " " ;}
    std::cout << std::endl;

    const std::vector<float> r2 =  CUDA_mult_MAT({7 , 8 , 9 , 10 , 11 , 12} , 3 , 2 ,
                                           {1 , 2 , 3 , 4 , 5 , 6} , 2 , 3);
/*
Product :
7  8   
9  10   1  2  3
11 12   4  5  6
*/


    for(auto & value: r2){std::cout << value << " " ;}
    std::cout << std::endl;

    return 0;
}

---

输出:

程序打印:

58 64 139 154 
39 54 69 49 68 87 0 0 0
                  ^~~~~~~

预期:

58 64 139 154 
39 54 69 49 68 87 59 82 105
                  ^~~~~~~

Answer 1

我们可以通过不同方式观察您的 CUBLAS 使用问题。

首先，研究CUBLAS Sgemm documentation ，我们看到 3 个参数 m、n、k 出现，按此顺序紧接在转置说明符之后:

cublasStatus_t cublasSgemm(cublasHandle_t handle,
                       cublasOperation_t transa, cublasOperation_t transb,
                       int m, int n, int k, 
                           ^      ^      ^

我们还观察到矩阵维度由下式给出:

A , B and C are matrices stored in column-major format with dimensions op ( A ) m × k , op ( B ) k × n and C m × n ,

所以第一个输入矩阵的维度是 m x k第二个输入矩阵的维度为 k x n，输出矩阵的维度为 m x n

让我们暂时关注输出矩阵。鉴于其维度是使用 m 和 n 参数指定的，它不可能是正确的(假设在非正方形的情况下)传递 only data_2 维度:

           cublasSgemm(handle , CUBLAS_OP_N , CUBLAS_OP_N,
                       data_2_columns , data_2_rows ,data_1_columns,
                       ^^^^^^^^^^^^^^   ^^^^^^^^^^^

其次，从错误检查的角度来看，通过使用 cuda-memcheck 运行您的代码，您可以快速估计您的 CUBLAS 调用有问题。报告的第一个错误如下:

$ cuda-memcheck ./t23
========= CUDA-MEMCHECK
========= Invalid __global__ read of size 4
=========     at 0x000006f0 in void gemmSN_NN_kernel<float, int=256, int=4, int=2, int=8, int=3, int=4, bool=0, cublasGemvTensorStridedBatched<float const >, cublasGemvTensorStridedBatched<float>>(cublasGemmSmallNParams<float const , cublasGemvTensorStridedBatched<float const >, float>)
=========     by thread (64,0,0) in block (0,0,0)
=========     Address 0x7f9c30a2061c is out of bounds
=========     Device Frame:void gemmSN_NN_kernel<float, int=256, int=4, int=2, int=8, int=3, int=4, bool=0, cublasGemvTensorStridedBatched<float const >, cublasGemvTensorStridedBatched<float>>(cublasGemmSmallNParams<float const , cublasGemvTensorStridedBatched<float const >, float>) (void gemmSN_NN_kernel<float, int=256, int=4, int=2, int=8, int=3, int=4, bool=0, cublasGemvTensorStridedBatched<float const >, cublasGemvTensorStridedBatched<float>>(cublasGemmSmallNParams<float const , cublasGemvTensorStridedBatched<float const >, float>) : 0x6f0)
=========     Saved host backtrace up to driver entry point at kernel launch time
=========     Host Frame:/usr/lib/x86_64-linux-gnu/libcuda.so.1 (cuLaunchKernel + 0x2b8) [0x1e5cc8]
=========     Host Frame:/usr/local/cuda/lib64/libcublasLt.so.11 [0x1063c8b]
=========     Host Frame:/usr/local/cuda/lib64/libcublasLt.so.11 [0x10a9965]
=========     Host Frame:/usr/local/cuda/lib64/libcublasLt.so.11 [0x6bfacc]
=========     Host Frame:/usr/local/cuda/lib64/libcublasLt.so.11 [0x5fc7af]
=========     Host Frame:/usr/local/cuda/lib64/libcublasLt.so.11 [0x436c35]
=========     Host Frame:/usr/local/cuda/lib64/libcublasLt.so.11 (cublasLtMatmul + 0x60f) [0x43484f]
=========     Host Frame:/usr/local/cuda/lib64/libcublas.so.11 [0x9ef6db]
=========     Host Frame:/usr/local/cuda/lib64/libcublas.so.11 [0x50e4f0]
=========     Host Frame:/usr/local/cuda/lib64/libcublas.so.11 (cublasSgemm_v2 + 0x1ee) [0x50f29e]
=========     Host Frame:./t23 [0x7986]
=========     Host Frame:./t23 [0x7b4c]
=========     Host Frame:/lib/x86_64-linux-gnu/libc.so.6 (__libc_start_main + 0xe7) [0x21b97]
=========     Host Frame:./t23 [0x744a]
=========

当然，一种可能的解决方案是转置输入矩阵，因此它们按列主要顺序排列，CUBLAS 提供了带有 Sgemm 的选项来执行此操作(见上文)。但是在我看来，您要做的是在不转置输入数组的情况下进行 C 风格的行主乘法。有文章here其中描述了如何做到这一点。

当我将该启发式应用于您的 cublasSgemm() 调用时，我得到了:

           cublasSgemm(handle , CUBLAS_OP_N , CUBLAS_OP_N,
                       data_2_columns , data_1_rows ,data_1_columns,
                       &alpha , GPU_data_2 , data_2_columns,
                       GPU_data_1 , data_1_columns,
                       &beta , GPU_result , data_2_columns)

当我使用这些更改编译并运行您的代码时，我得到:

$ cuda-memcheck ./t23
========= CUDA-MEMCHECK
58 64 139 154
39 54 69 49 68 87 59 82 105
========= ERROR SUMMARY: 0 errors