操作员<和严格的弱排序

问题描述

如何在 n 元组(例如在 3 元组上)定义 operator< 使其满足 严格弱排序 概念?我知道 boost 库具有正确定义的 operator< 元组类，但由于某些原因我不能使用它.

How to define operator< on n-tuple (for example on 3-tuple) so that it satisfy strict weak ordering concept ? I know that boost library has tuple class with correctly defined operator< but for some reasons I can't use it.

推荐答案

if (a1 < b1)
  return true;
if (b1 < a1)
  return false;

// a1==b1: continue with element 2
if (a2 < b2)
  return true;
if (b2 < a2)
  return false;

// a2 == b2: continue with element 3
if (a3 < b3)
  return true;
return false; // early out

这按照 a1 最重要和 a3 最不重要对元素进行排序.

This orders the elements by a1 being most siginificant and a3 least significant.

这可以无限期地继续，您也可以例如将其应用于 T 的向量，迭代 a[i] < 的比较.a[i+1]/a[i+1]

This can be continued ad infinitum, you could also e.g. apply it to a vector of T, iterating over comparisons of a[i] < a[i+1] / a[i+1] < a[i]. An alternate expression of the algorithm would be "skip while equal, then compare":

while (i<count-1 && !(a[i] < a[i+1]) && !(a[i+1] < a[i])
  ++i;
return i < count-1 && a[i] < a[i+1];

当然，如果比较的开销很大，您可能需要缓存比较结果.

Of course, if the comparison is expensive, you might want to cache the comparison result.

[edit] 删除了错误的代码

[edit] removed wrong code

[edit] 如果不仅仅是 operator< 可用，我倾向于使用模式

[edit] if more than just operator< is available, I tend to use the pattern

if (a1 != b1)
  return a1 < b1;

if (a2 != b2)
  return a2 < b2;

...

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