定义类的元素的词典比较的最简单的方法是什么？ [英] What's the simplest way of defining lexicographic comparison for elements of a class?

问题描述

如果我有一个类，我想能够排序（即支持一个小于概念），它有几个数据项，所以我需要做词典排序，然后我需要这样：

If I have a class that I want to be able to sort (ie support a less-than concept), and it has several data items such that I need to do lexicographic ordering then I need something like this:

struct MyData {
  string surname;
  string forename;

  bool operator<(const MyData& other) const {
    return surname < other.surname || (surname==other.surname && forename < other.forename); }
};

这对于超过2个数据成员的任何东西来说都变得难以管理。有什么更简单的方法来实现它吗？数据成员可以是任何Comparable类。

This becomes pretty unmanageable for anything with more than 2 data members. Are there any simpler ways of achieving it? The data members may be any Comparable class.

推荐答案

随着C ++ 11的出现，这使用 std :: tie ：

With the advent of C++11 there's a new and concise way to achieve this using std::tie:

bool operator<(const MyData& other) const {
  return std::tie(surname, forename) < std::tie(other.surname, other.forename);
}

这篇关于定义类的元素的词典比较的最简单的方法是什么？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！