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您好，我正在尝试运行一个程序，该程序使用蛮力和缓存技术（例如此处的 pdf）找到最接近的对：Caching Performance Stanford

我的原始代码是：

float compare_points_BF(int N,point *P){
    int i,j;
    float  distance=0, min_dist=FLT_MAX;
    point *p1, *p2;
    unsigned long long calc = 0;
    for (i=0;i<(N-1);i++){
        for (j=i+1;j<N;j++){
            if ((distance = (P[i].x - P[j].x) * (P[i].x - P[j].x) +
                    (P[i].y - P[j].y) * (P[i].y - P[j].y)) < min_dist){
            min_dist = distance;
            p1 = &P[i];
            p2 = &P[j];
            }
        }
    }
    return sqrt(min_dist);
}

这个程序给出了大约这些运行时间：

      N     8192    16384   32768   65536   131072  262144  524288  1048576      
 seconds    0,070   0,280   1,130   5,540   18,080  72,838  295,660 1220,576
            0,080   0,330   1,280   5,190   20,290  80,880  326,460 1318,631

上述程序的缓存版本为：

float compare_points_BF(register int N, register int B, point *P){
    register int i, j, ib, jb, num_blocks = (N + (B-1)) / B;
    register point *p1, *p2;
    register float distance=0, min_dist=FLT_MAX, regx, regy;

    //break array data in N/B blocks, ib is index for i cached block and jb is index for j strided cached block
    //each i block is compared with the j block, (which j block is always after the i block) 
    for (i = 0; i < num_blocks; i++){
        for (j = i; j < num_blocks; j++){
            //reads the moving frame block to compare with the i cached block
            for (jb = j * B; jb < ( ((j+1)*B) < N ? ((j+1)*B) : N); jb++){
                //avoid float comparisons that occur when i block = j block
                //Register Allocated
                regx = P[jb].x;
                regy = P[jb].y;
                for (i == j ? (ib = jb + 1) : (ib = i * B); ib < ( ((i+1)*B) < N ? ((i+1)*B) : N); ib++){
                    //calculate distance of current points
                    if((distance = (P[ib].x - regx) * (P[ib].x - regx) +
                            (P[ib].y - regy) * (P[ib].y - regy)) < min_dist){
                        min_dist = distance;
                        p1 = &P[ib];
                        p2 = &P[jb];
                    }
                }
            }
        }
    }
    return sqrt(min_dist);
}

和一些结果：

Block_size = 256        N = 8192        Run time: 0.090 sec
Block_size = 512        N = 8192        Run time: 0.090 sec
Block_size = 1024       N = 8192        Run time: 0.090 sec
Block_size = 2048       N = 8192        Run time: 0.100 sec
Block_size = 4096       N = 8192        Run time: 0.090 sec
Block_size = 8192       N = 8192        Run time: 0.090 sec


Block_size = 256        N = 16384       Run time: 0.357 sec
Block_size = 512        N = 16384       Run time: 0.353 sec
Block_size = 1024       N = 16384       Run time: 0.360 sec
Block_size = 2048       N = 16384       Run time: 0.360 sec
Block_size = 4096       N = 16384       Run time: 0.370 sec
Block_size = 8192       N = 16384       Run time: 0.350 sec
Block_size = 16384      N = 16384       Run time: 0.350 sec

Block_size = 128        N = 32768       Run time: 1.420 sec
Block_size = 256        N = 32768       Run time: 1.420 sec
Block_size = 512        N = 32768       Run time: 1.390 sec
Block_size = 1024       N = 32768       Run time: 1.410 sec
Block_size = 2048       N = 32768       Run time: 1.430 sec
Block_size = 4096       N = 32768       Run time: 1.430 sec
Block_size = 8192       N = 32768       Run time: 1.400 sec
Block_size = 16384      N = 32768       Run time: 1.380 sec

Block_size = 256        N = 65536       Run time: 5.760 sec
Block_size = 512        N = 65536       Run time: 5.790 sec
Block_size = 1024       N = 65536       Run time: 5.720 sec
Block_size = 2048       N = 65536       Run time: 5.720 sec
Block_size = 4096       N = 65536       Run time: 5.720 sec
Block_size = 8192       N = 65536       Run time: 5.530 sec
Block_size = 16384      N = 65536       Run time: 5.550 sec

Block_size = 256        N = 131072      Run time: 22.750 sec
Block_size = 512        N = 131072      Run time: 23.130 sec
Block_size = 1024       N = 131072      Run time: 22.810 sec
Block_size = 2048       N = 131072      Run time: 22.690 sec
Block_size = 4096       N = 131072      Run time: 22.710 sec
Block_size = 8192       N = 131072      Run time: 21.970 sec
Block_size = 16384      N = 131072      Run time: 22.010 sec

Block_size = 256        N = 262144      Run time: 90.220 sec
Block_size = 512        N = 262144      Run time: 92.140 sec
Block_size = 1024       N = 262144      Run time: 91.181 sec
Block_size = 2048       N = 262144      Run time: 90.681 sec
Block_size = 4096       N = 262144      Run time: 90.760 sec
Block_size = 8192       N = 262144      Run time: 87.660 sec
Block_size = 16384      N = 262144      Run time: 87.760 sec

Block_size = 256        N = 524288      Run time: 361.151 sec
Block_size = 512        N = 524288      Run time: 379.521 sec
Block_size = 1024       N = 524288      Run time: 379.801 sec

从我们可以看到运行时间比非缓存代码慢。这是由于编译器优化吗？代码是坏的还是仅仅是因为算法在平铺方面表现不佳？我使用用 32 位可执行文件编译的 VS 2010。提前致谢！

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2 回答 2

1

平铺可能是一个古老的概念，但它在今天仍然非常重要。在您的原始代码中，对于每个 i，您可以在仍然缓存的同时重用大部分 P[j] 元素，但前提是内部循环的长度足够小以适合那里。实际大小应由您要针对平铺的缓存级别确定 - L1 将提供最佳性能，因为它是最快的，但由于它也是最小的，您需要小块并且平铺开销可能太多。L2 允许更大的图块，但会略微降低性能，等等。

请注意，您不需要在这里使用 2d 平铺，这不是矩阵乘法 - 您正在遍历同一个数组。您可以简单地平铺内循环，因为它是溢出缓存的循环，一旦您这样做了 - 外循环 (i) 可以在当前缓存的内循环元素块上一直运行到最后。二维平铺实际上没有意义，因为没有人会重用外循环的元素（与矩阵 mul 相对）

因此，假设Point64 位大，您可以将 512 个这样的数组元素安全地放入 32k L1 中，或者将 4096 个元素放入 256k L2 中。如果 i 超出当前 j 块的范围，您将不得不错过每个块上的 P[i] 一次，但这可以忽略不计。

顺便说一句 - 这个解释可能仍然过时，因为一个足够好的编译器可能会尝试为你做这一切。虽然它相当复杂，所以我有点怀疑任何常见的甚至会尝试，但在这里应该很容易证明重新排序是安全的。当然，有人可能会争辩说“足够好的编译器”是一个悖论，但这是题外话......

于 2013-10-11T06:22:09.983 回答
1

这是一个有趣的案例。编译器在两个内部循环中的循环不变提升方面做得很差。即，两个内部 for 循环在每次迭代中检查以下条件：

(j+1)*B) < N ? ((j+1)*B) : N

和

(i+1)*B) < N ? ((i+1)*B) : N

计算和分支都很昂贵；但它们实际上对于两个内部 for 循环是循环不变的。一旦手动将它们从两个内部 for 循环中提升出来，我就能够使缓存优化版本的性能优于未优化版本（当 N==524288 时为 10%，当 N=1048576 时为 30%）。

这是修改后的代码（真的很简单，寻找u1，u2）：

//break array data in N/B blocks, ib is index for i cached block and jb is index for j strided cached block
//each i block is compared with the j block, (which j block is always after the i block) 
for (i = 0; i < num_blocks; i++){
    for (j = i; j < num_blocks; j++){
        int u1 =  (((j+1)*B) < N ? ((j+1)*B) : N);
        int u2 =  (((i+1)*B) < N ? ((i+1)*B) : N);
        //reads the moving frame block to compare with the i cached block
        for (jb = j * B; jb < u1 ; jb++){
            //avoid float comparisons that occur when i block = j block
            //Register Allocated
            regx = P[jb].x;
            regy = P[jb].y;
            for (i == j ? (ib = jb + 1) : (ib = i * B); ib < u2; ib++){
                //calculate distance of current points
                if((distance = (P[ib].x - regx) * (P[ib].x - regx) +
                        (P[ib].y - regy) * (P[ib].y - regy)) < min_dist){
                    min_dist = distance;
                    p1 = &P[ib];
                    p2 = &P[jb];
                }
            }
        }
    }
}
于 2013-10-11T21:03:30.170 回答