使用矩阵算法和常量进行嵌套 for 循环调试。

Question

这组嵌套 for 循环对于 M=64 和 N=64 的值可以正常工作，但当我使 M=128 和 N=64 时不起作用。我有另一个程序来检查矩阵乘法的正确值。直觉上它似乎仍然有效，但给了我错误的答案。

for(int m=64;m<=M;m+=64){
for(int n=64;n<=N;n+=64){
    for(int i = m-64; i < m; i+=16){

        float *A_column_start, *C_column_start;
        __m128 c_1, c_2, c_3, c_4, a_1, a_2, a_3, a_4, mul_1, 
               mul_2, mul_3, mul_4, b_1;
        int j, k;

        for(j = m-64; j < m; j++){

            //Load 16 contiguous column aligned elements from matrix C in
            //c_1-c_4 registers

            C_column_start = C+i+j*M;

            c_1 = _mm_loadu_ps(C_column_start);
            c_2 = _mm_loadu_ps(C_column_start+4);
            c_3 = _mm_loadu_ps(C_column_start+8);
            c_4 = _mm_loadu_ps(C_column_start+12);

            for (k=n-64; k < n; k+=2){

                //Load 16 contiguous column aligned elements from matrix A to
                //the a_1-a_4 registers

                A_column_start = A+k*M;

                a_1 = _mm_loadu_ps(A_column_start+i);
                a_2 = _mm_loadu_ps(A_column_start+i+4);
                a_3 = _mm_loadu_ps(A_column_start+i+8);
                a_4 = _mm_loadu_ps(A_column_start+i+12);

                //Load a value to resgister b_1 to act as a "B" or ("A^T") 
                //element to multiply against the A matrix

                b_1 = _mm_load1_ps(A_column_start+j);

                mul_1 = _mm_mul_ps(a_1, b_1);
                mul_2 = _mm_mul_ps(a_2, b_1);
                mul_3 = _mm_mul_ps(a_3, b_1);
                mul_4 = _mm_mul_ps(a_4, b_1);

                //Add together all values of the multiplied A and "B"
                //(or "A^T") matrix elements

                c_4 = _mm_add_ps(c_4, mul_4);
                c_3 = _mm_add_ps(c_3, mul_3);
                c_2 = _mm_add_ps(c_2, mul_2);
                c_1 = _mm_add_ps(c_1, mul_1);

                //Move over one column in A, and load the next 16 contiguous 
                //column aligned elements from matrix A to the a_1-a_4 registers

                A_column_start+=M;

                a_1 = _mm_loadu_ps(A_column_start+i);
                a_2 = _mm_loadu_ps(A_column_start+i+4);
                a_3 = _mm_loadu_ps(A_column_start+i+8);
                a_4 = _mm_loadu_ps(A_column_start+i+12);

                //Load a value to resgister b_1 to act as a "B" or "A^T"
                //element to multiply against the A matrix

                b_1 = _mm_load1_ps(A_column_start+j);

                mul_1 = _mm_mul_ps(a_1, b_1);
                mul_2 = _mm_mul_ps(a_2, b_1);
                mul_3 = _mm_mul_ps(a_3, b_1);
                mul_4 = _mm_mul_ps(a_4, b_1);

                //Add together all values of the multiplied A and "B" or
                //("A^T") matrix elements

                c_4 = _mm_add_ps(c_4, mul_4);
                c_3 = _mm_add_ps(c_3, mul_3);
                c_2 = _mm_add_ps(c_2, mul_2);
                c_1 = _mm_add_ps(c_1, mul_1);

            }
            //Store the added up C values back to memory

            _mm_storeu_ps(C_column_start, c_1);
            _mm_storeu_ps(C_column_start+4, c_2);
            _mm_storeu_ps(C_column_start+8, c_3);
            _mm_storeu_ps(C_column_start+12, c_4);

        }

    }
    }
}}

Answer 1

它适用于 M = 64 和 N=64，因为在这些情况下，您只在相应的循环（最外面的两个循环）中进行一次迭代。当 M = 128 时，您现在在外循环上执行两个步骤，在这种情况下，该行

C_column_start = C+i+j*M;

和该行将

A_column_start = A+k*M;

为内循环产生相同的结果，因此本质上对于您在外循环上执行的两个步骤（m=64,128）你只是将一个步骤的结果加倍，m=128。修复方法很简单，只需将 M 更改为 m 以便使用迭代变量即可。

此外，您还应该考虑对齐 A 和 C 中的数据，以便可以执行 SSE 对齐负载。这将导致更快的代码。

Answer 2

我猜您在代码中使用 M

C_column_start = C+i+j*M;

需要改为使用 m 。也可能在其他使用 M 的行中。
但是，我并不真正理解您的代码，因为您没有解释代码的用途，而且我不是数学程序员。