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改善局部性时的嵌套for循环范围(C ++)

[英]Ranges of nested for-loops when locality is improved (C++)

I have the following nested for loop: 我有以下嵌套的for循环:

int n = 8;
int counter = 0;

for (int i = 0; i < n; i++)
{
    for (int j = i + 1; j < n; j++)
    {
        printf("(%d, %d)\n", i, j);
        counter++;
    }
}

Which prints (0,1) to (6,7) as expected and the printf() statement is ran 28 times as indicated by counter . 如预期那样打印(0,1)至(6,7),并且printf()语句运行了28次,如counter

I have been the set the task of improving the efficiency of this code by improving its locality (this is test code, the value of n in the actual program is much larger and i and j are used to index into two 1d arrays) and have employed what I believe to be a fairly standard technique: 我已经承担了通过改善代码局部性来提高代码效率的任务(这是测试代码,实际程序中的n值要大得多,并且ij用于索引到两个1d数组中)使用了我认为是相当标准的技术:

int chunk = 4;

for(int i = 0; i < n; i+=chunk)
    for(int j = 0; j < n; j+=chunk)
        for (int i_chunk = 0; i_chunk < chunk; i_chunk++)
            for (int j_chunk = i_chunk + 1; j_chunk < chunk; j_chunk++)
            {
                printf("(%d, %d)\n", i+i_chunk, j+j_chunk);
                counter++;
            }

However, here printf() is only being ran 24 times because the j_chunk = i_chunk + 1 means that where before the j loop printed (0,1) to (0,7), the two iterations of the j_chunk loop where i+i_chunk == 0 print (0,1) to (0,3) and (0,5) to (0,7) missing (0,4). 但是,这里的printf()仅运行了24次,因为j_chunk = i_chunk + 1表示在j循环打印(0,1)到(0,7)之前, j_chunk循环的两次迭代是i+i_chunk == 0打印(0,1)至(0,3)和(0,5)至(0,7)缺少(0,4)。

I understand why it is doing this but I can't for the life of me come up with a solution; 我知道为什么这样做,但是我无法为自己的生活提出解决方案; any help would be appreciated. 任何帮助,将不胜感激。

First you need to make sure that j is never in a lower chunk than i , so your outer loops should be: 首先,您需要确保j永远不会比i低,因此您的外部循环应为:

for(int i = 0; i < n; i+=chunk)
   for(int j = i; j < n; j+=chunk)

Then you need different behaviour based on whether i and j are in the same chunk or not. 然后,根据ij是否在同一块中,您需要不同的行为。 If they are, j_chunk needs to allways be larger than i_chunk , otherwise you need to go through all possible combinations: 如果是,则j_chunk必须始终大于i_chunk ,否则,您需要进行所有可能的组合:

if(i==j)
{
    for (int i_chunk = 0; i_chunk < chunk; i_chunk++)
    {
        for (int j_chunk = i_chunk + 1; j_chunk < chunk; j_chunk++)
        {
            printf("(%d, %d)\n", i+i_chunk, j+j_chunk);
            counter++;
        }
    }
}
else
{
    for (int i_chunk = 0; i_chunk < chunk; i_chunk++)
    {
        for (int j_chunk = 0; j_chunk < chunk; j_chunk++)
        {
            printf("(%d, %d)\n", i+i_chunk, j+j_chunk);
            counter++;
        }
    }
}

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