用的std ::排序元素块排序（） [英] Sorting by blocks of elements with std::sort()
问题描述
我有边缘的数组，其定义为一个C数组双打，其中每4双打定义一个边缘，这样的：
I have an array of edges, which is defined as a C-style array of doubles, where every 4 doubles define an edge, like this:
double *p = ...;
printf("edge1: %lf %lf %lf %lf\n", p[0], p[1], p[2], p[3]);
printf("edge2: %lf %lf %lf %lf\n", p[4], p[5], p[6], p[7]);
所以我想用的std ::排序（）由边长排序。如果这是一个结构边缘{双X1，Y1，X2，Y2; };边缘* P; ，我会好到哪里去。
So I want to use std::sort() to sort it by edge length. If it was a struct Edge { double x1, y1, x2, y2; }; Edge *p;, I would be good to go.
但是，在这种情况下，双阵列具有块大小不是由指针型pssed前$ P $。 的qsort（）允许你明确指定块大小，但的std ::排序（） 推断块大小的指针类型。
But in this case, the double array has a block size that is not expressed by the pointer type. qsort() allows you to explicitly specify the block size, but std::sort() infers the block-size by the pointer type.
有关性能方面的原因（包括内存使用和CPU），让我们说，这是不可取的创建新阵列，或以某种方式变换的阵列。出于性能方面再次，让我们说，我们要使用的std ::排序（）而不是的qsort（）
For performance reasons (both memory-usage and CPU), let's say that it's undesirable to create new arrays, or transform the array somehow. For performance reasons again, let's say that we do want to use std::sort() instead of qsort().
是否有可能调用的std ::排序（）不会对转换数据浪费一个CPU周期？
Is it possible to call std::sort() without wasting a single CPU cycle on transforming the data?
这是显而易见的方法是尝试强制转换指针：
An obvious approach is to try to force-cast the pointer:
double *p = ...;
struct Edge { double arr[4]; };
Edge *p2 = reinterpret_cast<Edge*>(p);
std::sort(...);
但我怎么确保数据正确对齐？另外，我怎么确保它总是会在所有的平台和架构正确对齐？
But how do I make sure the data is aligned properly? Also, how do I make sure it will always be aligned properly on all platforms and architectures?
或者，我可以使用的typedef双[4]边缘;
推荐答案
怎么样有一个重新排序向量？您初始化1..1 / L载体，传递的std ::那种比较元素I1 * L..i1 * L + L到I2 * L..i2 * L + L，而当你的载体正确排序的比较，根据新的秩序重新排序C数组。
How about having a reordering vector? You initialize vector with 1..N/L, pass std::sort a comparator that compares elements i1*L..i1*L+L to i2*L..i2*L+L, and when your vector is properly sorted, reorder the C array according to new order.
在回应评论：是的事情变得复杂，但它可能只是好复杂！看看<一个href=\"http://stackoverflow.com/questions/838384/reorder-vector-using-a-vector-of-indices\">here.
In response to comment: yes things get complicated, but it may just be good complication! Take a look here.
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