“枚举类”的递增和递减 [英] Incrementation and decrementation of “enum class”

问题描述

据我们所知，C ++ 03中 enum 的递增和递减是非法的，因为C ++ 03 枚举可能不是连续的。但是，根据维基百科的说法，C ++ 11标准引入了新的枚举类，它更加类型安全，因为它不是基于任何简单的可数类型构建的。那么现在，如果我们有一个枚举值的有限列表，我们可以写一些类似

 枚举类颜色{黑色，蓝白色 }; 
 // ... 
颜色color = Colors :: White; 
 color ++; 
  

，它将正常工作（例如增加 White 将返回黑色，减少黑色将返回白色）？

如果我们无法编写这样的代码，你是否知道任何类似行为的类，从 boost 或 Qt 为我们提供了这个功能（正确的减少）？

解决方案

将正常工作

枚举不是按照您默认描述的方式环绕。

和C ++ 11的枚举类不保证连续的值，与C ++ 03的枚举相同。 p>

您可以为您的特定枚举定义包装行为。这个解决方案假设值是连续的，就像您描述的枚举一样。

 枚举类颜色{黑色，蓝色，白色，END_OF_LIST}; 
 
 // ++颜色的特殊行为
颜色& operator $（Colors& c）{
 using IntType = typename std :: underlying_type< Colors> :: type 
 c = static_cast< Colors>（static_cast< IntType>（c）+ 1）; 
 if（c == Colors :: END_OF_LIST）
 c = static_cast< Colors>（0）; 
 return c; 
} 
 
 //颜色的特殊行为++ 
颜色操作符++（颜色& c，int）{
颜色结果= c; 
 ++ c; 
返回结果; 
} 
  

As we know, incrementation and decrementation of enum in C++03 is illegal, because C++03 enum may not be continuous. But the C++11 standard introduced the new enum class construction, which, according to Wikipedia, is more type-safe because it isn’t built on any simple countable type. So now, if we have a bounded list of values of an enum, can we write something like

enum class Colors { Black, Blue, White };
// ...
Colors color = Colors::White;
color++;

and will it work correctly (e.g. incrementation of White will return Black and decrementation of Black will return White)?

If we can't write such code, do you know any behavior-like classes either from boost or from Qt that provide us this feature (correct in- and decrementing)?

解决方案

will it work correctly

No. enums are not designed to "wrap around" in the way you describe by default.

And C++11's enum class doesn't guarantee contiguous values, the same as you describe for C++03's enum.

You can define the wrapping behavior for your particular enum though. This solution assumes that the values are contiguous, like the enum you described.

enum class Colors { Black, Blue, White, END_OF_LIST };

// Special behavior for ++Colors
Colors& operator++( Colors &c ) {
  using IntType = typename std::underlying_type<Colors>::type
  c = static_cast<Colors>( static_cast<IntType>(c) + 1 );
  if ( c == Colors::END_OF_LIST )
    c = static_cast<Colors>(0);
  return c;
}

// Special behavior for Colors++
Colors operator++( Colors &c, int ) {
  Colors result = c;
  ++c;
  return result;
}

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