PDFJS和PDF编码(PDFJS and PDF encoding)

 我不确定如何有效地使用SIMD对任意矩阵进行就地转置，但我确实知道如何在不合适的地方进行。 让我来描述如何做到这两点 
 
 到位转置 
 
 对于就地转置，您应该在C ++手册中看到Agner Fog的优化软件 。 请参见第9.10节“大型数据结构中的缓存争用”示例9.5a。 对于某些矩阵大小，由于缓存别名，您将看到性能大幅下降。 有关示例，请参见表9.1， 为什么转换512x512的矩阵要比转置513x513的矩阵慢得多？ 。 Agner提供了一种方法来解决这个问题，使用示例9.5b中的循环平铺（类似于Paul R所描述的）。 
 
 不合适的转置 
 
 在这里查看我的答案（投票率最高的人） 在C ++中转换矩阵的最快方法是什么？ 。 我已经很久没看过了这个，但是让我在这里重复我的代码： 
 
inline void transpose4x4_SSE(float *A, float *B, const int lda, const int ldb) {
    __m128 row1 = _mm_load_ps(&A[0*lda]);
    __m128 row2 = _mm_load_ps(&A[1*lda]);
    __m128 row3 = _mm_load_ps(&A[2*lda]);
    __m128 row4 = _mm_load_ps(&A[3*lda]);
     _MM_TRANSPOSE4_PS(row1, row2, row3, row4);
     _mm_store_ps(&B[0*ldb], row1);
     _mm_store_ps(&B[1*ldb], row2);
     _mm_store_ps(&B[2*ldb], row3);
     _mm_store_ps(&B[3*ldb], row4);
}

inline void transpose_block_SSE4x4(float *A, float *B, const int n, const int m, const int lda, const int ldb ,const int block_size) {
    #pragma omp parallel for
    for(int i=0; i<n; i+=block_size) {
        for(int j=0; j<m; j+=block_size) {
            int max_i2 = i+block_size < n ? i + block_size : n;
            int max_j2 = j+block_size < m ? j + block_size : m;
            for(int i2=i; i2<max_i2; i2+=4) {
                for(int j2=j; j2<max_j2; j2+=4) {
                    transpose4x4_SSE(&A[i2*lda +j2], &B[j2*ldb + i2], lda, ldb);
                }
            }
        }
    }   
}

I'm not sure how to do a in-place transpose for arbitrary matrices using SIMD efficiently but I do know how to do it for out-of-place. Let me describe how to do both
 
In place transpose
 
For in-place transpose you should see Agner Fog's Optimizing software in C++ manual. See section 9.10 "Cache contentions in large data structures" example 9.5a. For certain matrix sizes you will see a large drop in performance due to cache aliasing. See table 9.1 for examples and this Why is transposing a matrix of 512x512 much slower than transposing a matrix of 513x513?. Agner gives a way to fix this using loop tiling (similar to what Paul R described) in Example 9.5b.
 
Out of place transpose
 
See my answer here (the one with the most votes) What is the fastest way to transpose a matrix in C++?. I have not looked into this in ages but let me just repeat my code here:
 
inline void transpose4x4_SSE(float *A, float *B, const int lda, const int ldb) {
    __m128 row1 = _mm_load_ps(&A[0*lda]);
    __m128 row2 = _mm_load_ps(&A[1*lda]);
    __m128 row3 = _mm_load_ps(&A[2*lda]);
    __m128 row4 = _mm_load_ps(&A[3*lda]);
     _MM_TRANSPOSE4_PS(row1, row2, row3, row4);
     _mm_store_ps(&B[0*ldb], row1);
     _mm_store_ps(&B[1*ldb], row2);
     _mm_store_ps(&B[2*ldb], row3);
     _mm_store_ps(&B[3*ldb], row4);
}

inline void transpose_block_SSE4x4(float *A, float *B, const int n, const int m, const int lda, const int ldb ,const int block_size) {
    #pragma omp parallel for
    for(int i=0; i<n; i+=block_size) {
        for(int j=0; j<m; j+=block_size) {
            int max_i2 = i+block_size < n ? i + block_size : n;
            int max_j2 = j+block_size < m ? j + block_size : m;
            for(int i2=i; i2<max_i2; i2+=4) {
                for(int j2=j; j2<max_j2; j2+=4) {
                    transpose4x4_SSE(&A[i2*lda +j2], &B[j2*ldb + i2], lda, ldb);
                }
            }
        }
    }   
}