允许基于范围对于枚举类？ [英] Allow for Range-Based For with enum classes?

问题描述

我有一个循环的代码块，其中我循环枚举类的所有成员。

我现在使用的循环与基于范围的新相比看起来很不实用。

有没有办法利用新的C ++ 11功能来减少我当前 for 循环的详细程度？ / p>

我想改进的当前代码：

 枚举类COLOR 
 {
 Blue，
 Red，
 Green，
 Purple，
 First = Blue，
 Last = Purple 
} 
 
 inline COLOR operator ++（COLOR& x）{return x =（COLOR）（（int）（x）+1））; } 
 
 int main（int argc，char ** argv）
 {
 //以任何方式改进基于范围的下一行？ 
 for（COLOR c = COLOR :: First; c！= COLOR :: Last; ++ c）
 {
 // do work 
} 
 return 0 ; 
} 
  

换句话说， / p>

  for（const auto& c：COLOR）
 {
 // do work 
} 
  

解决方案

枚举本身作为迭代器迭代枚举是糟糕的主意，我建议使用一个实际的迭代器，如deft_code的答案。但是如果这真的是你想要的：

  COLOR运算符++（COLOR& x）{return x =（COLOR） ：underlying_type< COLOR> :: type（x）+ 1）; } 
 COLOR操作符*（COLOR c）{return c;} 
 COLOR begin（COLOR r）{return COLOR :: First;} 
 COLOR end（COLOR r）{COLOR l = COLOR ：：持续;返回l ++;} 
 
 int main（）{
 for（const auto& c：COLOR（））{//注意我在这里添加了括号来创建一个实例
 // do work 
} 
 return 0; 
} 
  

在这里工作： http://ideone.com/cyTGD8

I have a recurrent chunk of code where I loop over all the members of an enum class.

The for loop that I currently use looks very unwieldly compared to the new range-based for.

Is there any way to take advantage of new C++11 features to cut down on the verbosity for my current for loop?

Current Code that I would like to improve:

enum class COLOR
{
    Blue,
    Red,
    Green,
    Purple,
    First=Blue,
    Last=Purple
};

inline COLOR operator++( COLOR& x ) { return x = (COLOR)(((int)(x) + 1)); }

int main(int argc, char** argv)
{
  // any way to improve the next line with range-based for?
  for( COLOR c=COLOR::First; c!=COLOR::Last; ++c )
  {
    // do work
  }
  return 0;
}

In other words, it would be nice if I could do something like:

for( const auto& c : COLOR )
{
  // do work
}

解决方案

Iterating enumerations with the enumeration itself as an iterator is a poor idea, and I recommend using an actual iterator as in deft_code's answer. But if this is really what you want:

COLOR operator++(COLOR& x) { return x = (COLOR)(std::underlying_type<COLOR>::type(x) + 1); }
COLOR operator*(COLOR c) {return c;} 
COLOR begin(COLOR r) {return COLOR::First;}
COLOR end(COLOR r)   {COLOR l=COLOR::Last; return l++;}

int main() { 
    for(const auto& c : COLOR()) { //note I added parenthesis here to make an instance
        //do work
    }
    return 0;
}

Working here: http://ideone.com/cyTGD8

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