3 回答 3

40

无强制转换的解决方案是使用 switch。但是,您可以使用模板生成伪开关。原理是使用模板列表(或参数包)递归处理枚举的所有值。所以,这是我找到的3种方法。

测试枚举:

enum class Fruit
{
    apple,
    banana,
    orange,
    pineapple,
    lemon
};

香草开关(住在这里)

Fruit& operator++(Fruit& f)
{
    switch(f)
    {
        case Fruit::apple:     return f = Fruit::banana;
        case Fruit::banana:    return f = Fruit::orange;
        case Fruit::orange:    return f = Fruit::pineapple;
        case Fruit::pineapple: return f = Fruit::lemon;
        case Fruit::lemon:     return f = Fruit::apple;
    }
}

C++03-ish 方法(住在这里)

template<typename E, E v>
struct EnumValue
{
    static const E value = v;
};

template<typename h, typename t>
struct StaticList
{
    typedef h head;
    typedef t tail;
};

template<typename list, typename first>
struct CyclicHead
{
    typedef typename list::head item;
};

template<typename first>
struct CyclicHead<void,first>
{
    typedef first item;
};

template<typename E, typename list, typename first = typename list::head>
struct Advance
{
    typedef typename list::head lh;
    typedef typename list::tail lt;
    typedef typename CyclicHead<lt, first>::item next;

    static void advance(E& value)
    {
        if(value == lh::value)
            value = next::value;
        else
            Advance<E, typename list::tail, first>::advance(value);
    }
};

template<typename E, typename f>
struct Advance<E,void,f>
{
    static void advance(E& value)
    {
    }
};

/// Scalable way, C++03-ish
typedef StaticList<EnumValue<Fruit,Fruit::apple>,
        StaticList<EnumValue<Fruit,Fruit::banana>,
        StaticList<EnumValue<Fruit,Fruit::orange>,
        StaticList<EnumValue<Fruit,Fruit::pineapple>,
        StaticList<EnumValue<Fruit,Fruit::lemon>,
        void
> > > > > Fruit_values;

Fruit& operator++(Fruit& f)
{
    Advance<Fruit, Fruit_values>::advance(f);
    return f;
}

C++11-ish 方法(住在这里)

template<typename E, E first, E head>
void advanceEnum(E& v)
{
    if(v == head)
        v = first;
}

template<typename E, E first, E head, E next, E... tail>
void advanceEnum(E& v)
{
    if(v == head)
        v = next;
    else
        advanceEnum<E,first,next,tail...>(v);
}

template<typename E, E first, E... values>
struct EnumValues
{
    static void advance(E& v)
    {
        advanceEnum<E, first, first, values...>(v);
    }
};

/// Scalable way, C++11-ish
typedef EnumValues<Fruit,
        Fruit::apple,
        Fruit::banana,
        Fruit::orange,
        Fruit::pineapple,
        Fruit::lemon
> Fruit_values11;

Fruit& operator++(Fruit& f)
{
    Fruit_values11::advance(f);
    return f;
}

(C++11-ish 旧版本)

您可以通过添加一些预处理器来扩展以消除重复值列表的需要。

于 2013-03-16T23:38:40.620 回答
5

C++ 中的每个枚举运算符都可以在不强制转换为基础类型的情况下编写,但结果会非常冗长。

举个例子:

size_t index( Colors c ) {
  switch(c) {
    case Colors::Black: return 0;
    case Colors::Blue: return 1;
    case Colors::White: return 2;
  }
}
Color indexd_color( size_t n ) {
  switch(n%3) {
    case 0: return Colors::Black;
    case 1: return Colors::Blue;
    case 2: return Colors::White;
  }
}
Colors increment( Colors c, size_t n = 1 ) {
  return indexed_color( index(c) + n );
}
Colors decrement( Colors c, size_t n = 1 ) {
  return indexed_color( index(c)+3 - (n%3) );
}
Colors& operator++( Colors& c ) {
  c = increment(c)
  return c;
}
Colors operator++( Colors& c, bool ) {
  Colors retval = c;
  c = increment(c)
  return retval;
}

并且智能编译器将能够将这些转换为直接在基本整数类型上的操作。

但是在您的接口中enum class转换为基本整数类型并不是一件坏事。运算符是您的enum class.

如果你不喜欢这个循环size_t并认为它是一个假演员,你可以写:

Colors increment( Colors c ) {
  switch(c) {
    case Colors::Black: return Colors::Blue;
    case Colors::Blue: return Colors::White;
    case Colors::White: return Colors::Black;
  }
}

和类似的递减,并实现递增n循环作为重复的循环increment

于 2013-03-16T17:22:04.393 回答
0
enum class Colors { Black, Blue, White };

Colors operator++(Colors& color)
{
    color = (color == Colors::White) ? Colors::Black : Colors(int(color) + 1);
    return color;
}

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于 2017-07-19T05:13:37.857 回答