矩阵转置的cache分析

最近在看一本关于C/C++性能优化的书,里面用矩阵转置说明了cpu cache使用效率问题,我也按照这个例子自己做了一个实验并进行了分析。完整的代码和说明也都放在 Github, 这里就把说明直接搬过来吧。

转置的核心代码非常简单,改进的方式就是不要试图把整行进行转置,而是对一个区域进行转置。对一个区域进行转置的好处是,按照跳行方式访问的那些行,在接下来还可以继续从cache里面访问到。

void Trans() {
    for(int i=1;i<N;i++) {
        for(int j=0;j<i;j++) {
            int x = data[i][j];
            data[i][j] = data[j][i];
            data[j][i] = x;
        }
    }
}

void FastTrans(const int stride) {
    for(int i=1;i<N;i+=stride) {
        int toi = min(N, i+stride);
        for(int j=0;j<i;j+=stride) {
            for (int ii=i;ii<toi;ii++) {
                int toj = min(ii, j+stride);
                for(int jj=j;jj<toj;jj++) {
                    int x = data[ii][jj];
                    data[ii][jj] = data[jj][ii];
                    data[jj][ii] = x;
                }
            }
        }
    }
}

下面是运行结果,其中FAST是使用 stride=8 跑出来的

YAN007 :: ~/shared/MatrixTranspose ‹master*› » ./a.out
[SLOW, S=8] N = 1010, took: 1533ms, avg 1502.79ns/N
[FAST, S=8] N = 1010, took: 1582ms, avg 1550.83ns/N
[SLOW, S=8] N = 1024, took: 10529ms, avg 10041.2ns/N
[FAST, S=8] N = 1024, took: 2382ms, avg 2271.65ns/N
[SLOW, S=8] N = 1030, took: 1628ms, avg 1534.55ns/N
[FAST, S=8] N = 1030, took: 1467ms, avg 1382.79ns/N

此时stride设置是8,可以看到效果还是不错的,对于N=1030有所改进,对于N=1024来说提升就特别大。

如果cache size = 64KB, ways = 4, line size = 64, 矩阵是1024 * 1024的话,那么:

  1. 相邻两个row, 地址差距是 1024 * 4 = 4kB
  2. cache size = 64KB, line size = 64, ways = 4, 那么一共有256个slots
  3. 所以理论上两个row都会落在一个slot上,但是因为ways=4, 所以可以忍受4个冲突。

不过如果是使用SLOW的方法,cache是没有办法忍受冲突的,因此是按照行扫描下来的,然后继续从0行开始扫描,而此时缓存已经全部被换出了。

在Linux下面可以查看cache的配置

YAN007 :: ~/shared/MatrixTranspose ‹master*› » getconf -a | grep CACHE
LEVEL1_ICACHE_SIZE                 32768
LEVEL1_ICACHE_ASSOC                8
LEVEL1_ICACHE_LINESIZE             64
LEVEL1_DCACHE_SIZE                 32768
LEVEL1_DCACHE_ASSOC                8
LEVEL1_DCACHE_LINESIZE             64
LEVEL2_CACHE_SIZE                  1048576
LEVEL2_CACHE_ASSOC                 16
LEVEL2_CACHE_LINESIZE              64
LEVEL3_CACHE_SIZE                  20185088
LEVEL3_CACHE_ASSOC                 11
LEVEL3_CACHE_LINESIZE              64
LEVEL4_CACHE_SIZE                  0
LEVEL4_CACHE_ASSOC                 0
LEVEL4_CACHE_LINESIZE              0

这里看L1D和L2配置,好像L3的效果就不明显了。

处理1024*1024的矩阵,从L1D上看最多承受8个冲突,从L2上看最多承受64个冲突。然后因为line size = 64B, 最多可以放入16个int. 如果我们想提高L1D的cache命中率,stride就不能超过8个,考虑到按照行本身需要占用1个cache line, 所以stride估计设置成为7个是最好的。

下面是 stride=7 的效果

YAN007 :: ~/shared/MatrixTranspose ‹master*› » ./a.out
[SLOW, S=7] N = 1010, took: 1513ms, avg 1483.19ns/N
[FAST, S=7] N = 1010, took: 1637ms, avg 1604.74ns/N
[SLOW, S=7] N = 1024, took: 10520ms, avg 10032.7ns/N
[FAST, S=7] N = 1024, took: 2339ms, avg 2230.64ns/N
[SLOW, S=7] N = 1030, took: 1627ms, avg 1533.6ns/N
[FAST, S=7] N = 1030, took: 1490ms, avg 1404.47ns/N

下面是 stride=6 的效果

YAN007 :: ~/shared/MatrixTranspose ‹master*› » ./a.out
[SLOW, S=6] N = 1010, took: 1528ms, avg 1497.89ns/N
[FAST, S=6] N = 1010, took: 1620ms, avg 1588.08ns/N
[SLOW, S=6] N = 1024, took: 10528ms, avg 10040.3ns/N
[FAST, S=6] N = 1024, took: 2398ms, avg 2286.91ns/N
[SLOW, S=6] N = 1030, took: 1635ms, avg 1541.14ns/N
[FAST, S=6] N = 1030, took: 1507ms, avg 1420.49ns/N

下面是 stride=9 的效果,这个估计会稍微查一些,因为在处理一个stride的时候回产生 stride 次cache miss

YAN007 :: ~/shared/MatrixTranspose ‹master*› » ./a.out
[SLOW, S=9] N = 1010, took: 1514ms, avg 1484.17ns/N
[FAST, S=9] N = 1010, took: 1531ms, avg 1500.83ns/N
[SLOW, S=9] N = 1024, took: 10612ms, avg 10120.4ns/N
[FAST, S=9] N = 1024, took: 3489ms, avg 3327.37ns/N
[SLOW, S=9] N = 1030, took: 1653ms, avg 1558.11ns/N
[FAST, S=9] N = 1030, took: 1450ms, avg 1366.76ns/N