从两个枚举类创建复合类型，准备好STL映射 [英] Creating a composite type from two enum classes, ready for STL map

问题描述

我想要从两个枚举类中创建一个复合类型。

I would like to create a composite type out of two enum classes.

enum class Color {RED, GREEN, BLUE};
enum class Shape {SQUARE, CIRCLE, TRIANGLE};

class Object {
  Color color;
  Shape shape;
public:
};

为了在STL中使用 Object container like std :: map<> 我需要重载less-than操作符。然而，为了将两个枚举类变成一个线性索引，我需要枚举类的元素数量（NoE）：

In order to use Object in an STL container like std::map<> I would need to overload the less-than operator. However, in order to flatten both enum classes into one linear index I somehow need the number of elements (NoE) of the enum classes:

friend bool operator< (const Object &lhs, const Object &rhs) {
  return NoE(Shape)*lhs.color+lhs.shape < NoE(Shape)*rhs.color+rhs.shape;
}

如果不输入相同的信息两个地方在节目中以一种很好的方式？ （好的方式意味着没有 FIRST_ELEMENT，LAST_ELEMENT ，预处理魔法等）

How can this be done without entering the same information (number of elements) in two places in the program in a nice way? (Nice way means no FIRST_ELEMENT, LAST_ELEMENT, preprocessor magic, etc.)

问题（枚举中的元素数量）类似，但不涉及枚举类。

Question (Number of elements in an enum) is similar but does not address enum classes.

我想知道什么是在C ++ 11中实现这种类型的复合类型的最好方法。是枚举类定义足够强还是有必要说：？

I would like to know what is the best way to implement this kind of composite types in C++11. Is the enum class definition strong enough, or is it necessary to say:?

enum class Color {RED=0, GREEN=1, BLUE=2};
enum class Shape {SQUARE=0, CIRCLE=1, TRIANGLE=2};

推荐答案

如其他人所说，到运算符中的形状或颜色比较其他如果第一个相等。
使用执行运算符的替代方法 std :: tie ：

As commented and as already stated by others, give precedence to either Shape or Color in the operator< and only compare the other if the first is equal. An alternative implementation for operator< using std::tie:

#include <tuple>
friend bool operator<(const Object& lhs, const Object& rhs)
{
    return std::tie(lhs.color, lhs.shape) < std::tie(rhs.color, rhs.shape);
}

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