本文主要是介绍leading dimension，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
矩阵空间是 3x4，其左上角有一个子矩阵2x3，表示如下
11 22 33 0
44 55 66 0
0 0 0 0
i， j分别表示行索引，列索引
如果用列存储的话，leading dimension = 3(矩阵空间的行个数)， 换算公式是i + j *ld
11 44 0 22 55 0 33 66 0 0 0 0
如果是用行存储， leading dimension = 4(矩阵空间的列个数)，换算公式是 i*ld + j
11 22 33 0 44 55 66 0 0 0 0 0
cublas中矩阵用列存储表示，cusparse中的矩阵用行CNC表示
在cuda中二维数组是按照列存储的，在c中二维数组按照行存储，在c语言中的一个矩阵（二维数组）A = M X K, B = N X K, C = A * Bt = M x N
对于cuda而言（列存储），看到的A矩阵是K x M, B 是 K x N, 计算的C = Bt * A = N x M
计算结果C矩阵在c语言看来就是按照行存储的 M x N
在cuda中，对于A矩阵，不论在cublasSgemm, 是trans还是non-trans，其leading dimension就是 K, 同理对于矩阵B，是转置还是不转置，leading dimension都是K
在cuda中
设 A' = B = K x N , B' = A = K x M
transa = CUBLAS_OP_T
transb = CUBLAS_OP_N
m = op(A')_row = N
n = op(B')_col = M
k = op(A')_col = op(B')_row = K
lda = A'_row = K
ldb = B'_row = K
ldc = C_row = N
在c语言中有上图的矩阵，矩阵空间是 M x N, 但是只用到了mxn的子矩阵
对cuda而言，它看到的矩阵空间是NxM, 子矩阵是nxm
调用cublasSgemm，m,n,k都是用子矩阵的大小维度，但是lda = N
总结就是，gemm中的n,m,k都是计算的在cuda视角下子矩阵的维度，随着矩阵转置与否变化，但是lda是在cuda视角下矩阵空间的row的大小，并且不随矩阵转置与否变化
参考http://stackoverflow.com/questions/14595750/transpose-matrix-multiplication-in-cublas-howto
The problem is simple: I have two matrices, A and B, that are M by N, where M >> N. I want to first take the transpose of A, and then multiply that by B (A^T * B) to put that into C, which is N by N. I have everything set up for A and B, but how do I call cublasSgemm properly without it returning the wrong answer?
I understand that cuBlas has a cublasOperation_t enum for transposing things beforehand, but somehow I'm not quite using it correctly. My matrices A and B are in row-major order, i.e. [ row1 ][ row2 ][ row3 ]..... in device memory. That means for A to be interpreted as A-transposed, BLAS needs to know my A is in column-major order. My current code looks like below:
float *A, *B, *C;
// initialize A, B, C as device arrays, fill them with values
// initialize m = num_row_A, n = num_row_B, and k = num_col_A;
// set lda = m, ldb = k, ldc = m;
// alpha = 1, beta = 0;
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