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archibate
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Jan 29, 2022
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感谢第一个提交作业！
- 完成作业基本要求 42/50 分
- 能够在 ANSWER.md 中用自己的话解释 23/25 分
- 代码格式规范、能够跨平台 4/5 分
- 有自己独特的创新点 11/20 分
| // 这两个是临时变量，有什么可以优化的？ 5 分 | ||
| Matrix Rt, RtA; | ||
| // ans: 改为static变量，预先分配好空间。 | ||
| static Matrix Rt(std::array<std::size_t, 2>{1024, 1024}), RtA(std::array<std::size_t, 2>{1024, 1024}); |
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| static Matrix Rt(std::array<std::size_t, 2>{1024, 1024}), RtA(std::array<std::size_t, 2>{1024, 1024}); | |
| static thread_local Matrix Rt, RtA; |
我觉得可以一开始为空没问题。thread_local保证如果多个线程访问不会出错。
| for(int i=0; i<nx; i+=32){ | ||
| for(int t=0; t<nt; t++){ | ||
| for(int i_block=i; i_block<i+32; i_block++){ | ||
| out(i,j) += lhs(i_block, t) * rhs(t, j); |
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| out(i,j) += lhs(i_block, t) * rhs(t, j); | |
| out(i_block,j) += lhs(i_block, t) * rhs(t, j); |
漏改了一个？
| for (int y = 0; y < ny; y++) { | ||
| float val = wangsrng(x, y).next_float(); | ||
| out(x, y) = val; | ||
| // ans: 矩阵的x轴是紧密排列的，但是循环的内循环是y，访问数据时会跳跃，不利于cache; |
| out(y, x) = in(x, y); | ||
| } | ||
| } | ||
| // ans: 因为out矩阵是紧密访问，但是in矩阵是跳跃访问，cache中放不下。应改为分块转置。 |
| out(x, y) = 0; // 有没有必要手动初始化？ 5 分 | ||
| for (int t = 0; t < nt; t++) { | ||
| out(x, y) += lhs(x, t) * rhs(t, y); | ||
| // ans: lhs是跳跃访问，rhs是连续访问，out不动，造成无法矢量化。 |
| TICK(matrix_RtAR); | ||
| // 这两个是临时变量，有什么可以优化的？ 5 分 | ||
| Matrix Rt, RtA; | ||
| // ans: 改为static变量，预先分配好空间。 |
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3，应该加thread_local，不需要初始化大小。
| // #pragma omp parallel for collapse(2) | ||
| // for (int y = 0; y < ny; y++) { | ||
| // for (int x = 0; x < nx; x++) { | ||
| // out(x, y) = 0; // 有没有必要手动初始化？ 5 分 |
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请见ANSWER.md。