我有一个大小为10x20的输入矩阵A，我想按如下方式对其列进行置换：
p=[1 4 2 3 5 11 7 13 6 12 8 14 17 9 15 18 10 16 19 20] ;%rearrange the columns of A
A=A(:,p);
为此，我构造了一个对应于置换向量p的置换矩阵I，并且可以通过执行以下乘法来获得置换A：
A=A*I
我在Matlab中测试了排列，一切都很好。现在，我想使用cublas在cuda中进行测试。
输入矩阵A在列major中输入。列专业中的置换矩阵I也是如此。以下代码仅用于测试排列：
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <cublas_v2.h>
#define cudacall(call) \
do \
{ \
cudaError_t err = (call); \
if(cudaSuccess != err) \
{ \
fprintf(stderr,"CUDA Error:\nFile = %s\nLine = %d\nReason = %s\n", __FILE__, __LINE__, cudaGetErrorString(err)); \
cudaDeviceReset(); \
exit(EXIT_FAILURE); \
} \
} \
while (0)
#define cublascall(call) \
do \
{ \
cublasStatus_t status = (call); \
if(CUBLAS_STATUS_SUCCESS != status) \
{ \
fprintf(stderr,"CUBLAS Error:\nFile = %s\nLine = %d\nCode = %d\n", __FILE__, __LINE__, status); \
cudaDeviceReset(); \
exit(EXIT_FAILURE); \
} \
\
} \
while(0)
__global__ void sgemm_kernel(float *A_d, float *I_d)
{
int m=10,n=20,k=20;
int lda=k, ldb=k;
cublasHandle_t hdl;
cublasStatus_t status = cublasCreate_v2(&hdl);
const float alpha=1.0F, beta=0.0f;
status=cublasSgemm(hdl,CUBLAS_OP_N,CUBLAS_OP_N,k,n,k,&alpha,A_d,lda,I_d,ldb,&beta,A_d,lda);
}
int main(int argc, char* argv[])
{float A[10*20]={-0.0614, -0.0199, 0.0024, -0.0414, 0.1736, -0.0595, -0.2794, 0.1946, -0.0647, -0.0025,
-0.0036, 0.0628, -0.0827, 0.3679, -0.1913, 0.0500, -0.0245, 0.3855, -0.1298, -0.0334,
-0.0241, -0.0564, 0.0098, -0.2862, -0.0474, 0.0333, -0.3049, 0.2851, -0.1242, 0.0162,
0.0241, 0.0270, -0.0670, 0.3129, -0.2428, 0.0947, -0.1878, 0.0889, -0.0208, 0.0075,
-0.1559, 0.1437, -0.1916, 0.2297, -0.0833, -0.1805, 0.2522, -0.1738, 0.1027, -0.1273,
0.0716, 0.1882, -0.0963, 0.1081, 0.0958, -0.0713, 0.1931, 0.0874, -0.4186, 0.0345,
-0.1912, 0.0501, -0.1396, -0.0989, -0.0338, 0.1773, 0.1088, 0.0389, -0.0117, 0.0014,
0.1648, -0.1705, -0.0575, -0.0133, -0.0570, 0.2124, -0.0193, 0.1535, 0.0857, -0.1308,
0.1971, 0.0882, -0.2577, 0.1662, -0.2498, -0.0365, -0.1805, 0.0921, 0.0912, 0.0178,
-0.0379, 0.0080, 0.0572, -0.0067, 0.0591, -0.0136, 0.0471, -0.0163, 0.0082, -0.0338,
-0.2436, 0.1116, 0.0732, -0.0319, 0.0550, 0.2821, 0.0240, 0.0109, -0.0034, 0.1212,
-0.0061, 0.2497, -0.0542, -0.0939, 0.0651, 0.0063, -0.1367, 0.0580, 0.7389, -0.1143,
-0.3786, 0.1288, 0.0001, 0.2604, -0.1094, -0.3624, -0.0184, 0.0538, 0.0329, 0.0040,
0.0603, 0.1422, 0.1037, -0.1846, 0.4046, -0.3738, -0.3487, 0.3846, -0.0849, 0.0135,
-0.1850, 0.3571, -0.0543, -0.0025, -0.2880, 0.0600, 0.2605, -0.0474, 0.0010, -0.0333,
-0.1974, 0.4788, -0.2441, 0.3847, -0.1235, -0.3503, -0.1785, -0.1095, 0.3158, 0.0062,
-0.0509, -0.0502, 0.2154, 0.2237, -0.0671, 0.0377, 0.0519, 0.1530, -0.1675, 0.1856,
-0.0380, -0.0026, 0.4700, 0.0097, -0.2394, 0.0717, -0.2101, 0.2841, -0.1799, -0.0924,
-0.2678, 0.4485, 0.0044, 0.0030, -0.0439, 0.4337, 0.1819, -0.0180, -0.5443, 0.0864,
0.0390, -0.0235, -0.0706, 0.0138, 0.0633, -0.0147, 0.0444, -0.0334, 0.0557, 0.0507}
float I[20*20]={1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1};
float *A_d, *I_d;
cudacall(cudaMalloc(&A_d,10*20*sizeof( float )));
cudacall(cudaMalloc(&I_d, 20*20*sizeof(float )));
cudacall(cudaMemcpy(A_d, A, 10*20*sizeof(float), cudaMemcpyHostToDevice));
cudacall(cudaMemcpy(I_d, I, 20*20*sizeof(float), cudaMemcpyHostToDevice));
sgemm_kernel<<<1,1>>>(A_d, I_d);
cudacall(cudaDeviceSynchronize());
cudacall(cudaMemcpy(A, A_d, 10*20*sizeof(float), cudaMemcpyDeviceToHost));
cudacall(cudaFree(A_d));
cudacall(cudaFree(I_d));
return 0;
}
我无法得到正确的结果。
答案 0 :(得分：0)
CUBLAS不支持就地操作（事实上我没有支持它的并行BLAS）。您不能传递A_d并在乘法中使用它并将其作为操作中的矩阵使用。您必须使用不同的内存分配来保存结果。
所以
C <- 1*(A * B) + 0*C
是合法的，而
A <- 1*(A * B) + 0*A
不是。
答案 1 :(得分：-1)
cublasSgemm是一个宿主函数，因此应该从没有__global__限定符的函数中调用它。