允许基于范围的 For 与枚举类?

Allow for Range-Based For with enum classes?(允许基于范围的 For 与枚举类?)

问题描述

我有一个循环代码块，其中我循环了 enum 类 的所有成员.

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

与新的 基于范围的 for 相比，我目前使用的 for 循环看起来非常笨拙.

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

有什么方法可以利用 C++11 的新特性来减少当前 for 循环的冗长?

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

我想改进的当前代码:

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;
}

换句话说，如果我能做这样的事情就好了:

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

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

推荐答案

用枚举本身作为迭代器来迭代枚举是一个糟糕的主意，我建议使用实际的迭代器，如 deft_code 的答案.但如果这真的是你想要的:

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() { 
    //note the parenthesis after COLOR to make an instance
    for(const auto& c : COLOR()) {
        //do work
    }
    return 0;
}

在这里工作:http://ideone.com/cyTGD8

const COLOR COLORS[] = {COLOR::Blue, COLOR::Red, COLOR::Green, COLOR::Purple};
const COLOR (&COLORREF)[(int)COLOR::Last+1] = COLORS;

int main() { 
    for(const auto& c : COLORS) {
        //do work
    }
    return 0;
}

如下所示:http://coliru.stacked-crooked.com/a/5d356cc91556d6ef

(如果颜色数量与数组中的元素数量不匹配，单独的定义和数组的引用会导致编译器错误.非常容易安全检查.)

(The separate defintinion and the reference of the array makes it a compiler error if the number of colors doesn't match the number of elements in the array. Excellent easy safety check.)

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