ADA亚型等效于C

no，不是本地。

您所描述的可能是最好的表示为范围的枚举，并伴随着单独的范围枚举，并带有一组枚举的子集，这些枚举与"父母"范围示波器共享数值表示。

您可以进一步定义两者之间的一些转换，但是如果没有反思，就不可能使其全部优雅和直观，至少不是没有硬编码和重复的东西，而不是打败了目的。

在编程C 时，最好尝试完全放弃通过其他语言编程所填写的思维方式。

话虽如此，这实际上是一个很好的功能主意，尽管我不会屏住呼吸！

解决方法：只需使用枚举，然后应用范围检查您需要的位置。

C 枚举和ADA枚举之间还有一些其他差异。以下ADA代码演示了其中一些差异。

with Ada.Text_IO; use Ada.Text_IO;
procedure Subtype_Example is
   type Days is (Monday, Tueday, Wednesday, Thursday, Friday, Saturday, Sunday);
   subtype Work_Days is Days range Monday..Friday;
begin
   Put_Line("Days of the week:");
   for D in Days'Range loop
      Put_Line(D'Image);
   end loop;
   New_Line;
   Put_Line("Days with classification:");
   for D in Days'Range loop
      Put(D'Image & " is a member of");
      if D in Work_Days then
         Put_Line(" Work_Days");
      else
         Put_Line(" a non-work day");
      end if;
   end loop;
end Subtype_Example;

该程序的输出是：

Days of the week:
MONDAY
TUEDAY
WEDNESDAY
THURSDAY
FRIDAY
SATURDAY
SUNDAY
Days with classification:
MONDAY is a member of Work_Days
TUEDAY is a member of Work_Days
WEDNESDAY is a member of Work_Days
THURSDAY is a member of Work_Days
FRIDAY is a member of Work_Days
SATURDAY is a member of a non-work day
SUNDAY is a member of a non-work day

subtype work_days与类型天有IS-A关系。工作_DAYS的每个成员也是日子的成员。在此示例中

ADA中的字符定义为枚举。因此，为特殊用途定义类型字符的子类型很容易。以下示例读取文件中的文本，并计算上大写字母和下案字母的出现数量，忽略了文件中的所有其他字符。

with Ada.Text_IO; use Ada.Text_IO;
procedure Count_Letters is
   subtype Upper_Case is Character range 'A'..'Z';
   subtype Lower_Case is Character range 'a'..'z';
   Uppers : array(Upper_Case) of Natural;
   Lowers : array(Lower_Case) of Natural;
   File_Name : String(1..1024);
   File_Id   : File_Type;
   Length    : Natural;
   Line      : String(1..100);
begin
   -- set the count arrays to zero
   Uppers := (Others => 0);
   Lowers := (Others => 0);
   Put("Enter the name of the file to read: ");
   Get_Line(Item => File_Name,
            Last => Length);
   -- Open the named file
   Open(File => File_Id,
        Mode => In_File,
        Name => File_Name(1..Length));
   -- Read the file one line at a time
   while not End_Of_File(File_Id) loop
      Get_Line(File => File_Id,
               Item => Line,
               Last => Length);
      -- Count the letters in the line
      for I in 1..Length loop
         if Line(I) in Upper_Case then
            Uppers(Line(I)) := Uppers(Line(I)) + 1;
         elsif Line(I) in Lower_Case then
            Lowers(Line(I)) := Lowers(Line(I)) + 1;
         end if;
      end loop;
   end loop;
   Close(File_Id);
   -- Print the counts of upper case letters
   for Letter in Uppers'Range loop
      Put_Line(Letter'Image & " =>" & Natural'Image(Uppers(Letter)));
   end loop;
   -- print the counts of lower case letters
   for Letter in Lowers'Range loop
      Put_Line(Letter'Image & " =>" & Natural'Image(Lowers(Letter)));
   end loop;
end Count_Letters;

定义了两个字符的亚型。subtype upper_case包含字符值范围从" a"到" z"，而subtype lower_case包含字符值范围从" a"到't t os'z'。

创建了两个数组来计算读取的字母。数组鞋面由upper_case值集索引。数组的每个元素都是天然的实例，它是仅包含非负值的整数的预定义亚型。较低的较低_case值索引阵列索引。降低的每个元素也是自然的一个实例。

程序提示文件名，打开该文件，然后一次读取文件一行。每行中的字符被解析。如果字符是上_case字符，则额定字母索引中的数组元素会增加。如果字符是person_case字符，则降低字母索引的数组元素会增加。

以下输出是读取count_letters程序的源文件的结果。

Enter the name of the file to read: count_letters.adb
'A' => 3
'B' => 0
'C' => 12
'D' => 0
'E' => 2
'F' => 13
'G' => 2
'H' => 0
'I' => 21
'J' => 0
'K' => 0
'L' => 36
'M' => 1
'N' => 9
'O' => 7
'P' => 4
'Q' => 0
'R' => 3
'S' => 2
'T' => 3
'U' => 9
'V' => 0
'W' => 0
'X' => 0
'Y' => 0
'Z' => 1
'a' => 51
'b' => 3
'c' => 8
'd' => 19
'e' => 146
'f' => 15
'g' => 16
'h' => 22
'i' => 50
'j' => 0
'k' => 0
'l' => 38
'm' => 13
'n' => 57
'o' => 48
'p' => 35
'q' => 0
'r' => 62
's' => 41
't' => 78
'u' => 19
'v' => 0
'w' => 12
'x' => 2
'y' => 6
'z' => 2

使用C 17引入的std::variant实现您想要的可能(至少部分(。

struct Monday {};
struct Tuesday {};
/* ... etc. */
using WeekDay= std::variant<Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday>;

以下代码定义了sub_variant_t，该代码从提交的类型中构造了新的variant。例如。using Working_Day= sub_variant_t<WeekDay,5>;从Weekday获取前五个元素。

template<class T,size_t o,class S>
struct sub_variant_h;
template<class T,size_t o,size_t... I>
struct sub_variant_h<T,o,std::index_sequence<I...> >
{
    using type= std::variant<typename std::variant_alternative_t<(I+o),T>... >;
};
template<class T,size_t end, size_t beg=0>
struct sub_variant
{
    using type= typename sub_variant_h<T,beg,std::make_index_sequence<end-beg> >:type;
};
template<class T,size_t end, size_t beg=0>
using sub_variant_t = typename sub_variant<T,end,beg>::type;

如果要从较小 type( Working_Day(复制值较大一个(Weekday(，则可以使用WeekDay d3= var2var<WeekDay>( d1 );，其中var2var定义为如下。

template<class toT, class... Types>
toT
var2var( std::variant<Types...> const & v )
{
    return std::visit([](auto&& arg) -> toT {return toT(arg);}, v);
}

请参阅此Livedemo。

可能您可以将分配重载与后条件

Ensures(result > 0 && result < 10);  

纯粹的理论。没有尝试过自己。但是你们怎么看？

，但是很有趣地欣赏他们如何将C 上的每次升级为高级功能 ADA程序员认为是理所当然的所有内容。

范围检查的成本。C 的功能具有零成本策略：如果您想要该功能，并且应该为其支付成本，则需要明确。话虽如此，大多数情况下您可以使用一些图书馆或编写自己的图书馆。

另外，当某人试图将Sunday放入Working_Day时，您期望什么？例外(很可能(？将其设置为Monday？将其设置为Friday？使对象无效？保持相同的价值并忽略(坏主意(？

作为一个例子：

#include <iostream>
#include <string>
using namespace std;
enum class Weekday
{
    Sunday= 0,
    Monday,
    Tuesday,
    Wednesday,
    Thursday,
    Friday,
    Saturday
};
template <class T, T min, T max>
class RangedAccess
{
    static_assert(max >= min, "Error min > max");
private:
    T t;
    public:
    RangedAccess(const T& value= min)
    {
        *this= value;
    }

    RangedAccess& operator=(const T& newValue)
    {
        if (newValue > max || newValue < min) {
            throw string("Out of range");
        }
        t= newValue;
    }
    operator const T& () const
    { 
        return t; 
    }
    const T& get() const
    { 
        return t; 
    }
};
using Working_Day= RangedAccess<Weekday, Weekday::Monday, Weekday::Friday>;
int main()
{
    Working_Day workday;
    cout << static_cast<int>(workday.get()) << endl; // Prints 1
    try {
        workday= Weekday::Tuesday;
        cout << static_cast<int>(workday.get()) << endl; // Prints 2
        workday= Weekday::Sunday; // Tries to assign Sunday (0), throws
        cout << static_cast<int>(workday.get()) << endl; // Never gets executed
    } catch (string s) {
        cout << "Exception " << s << endl; // Prints "Exception out of range"
    }
    cout << static_cast<int>(workday.get()) << endl; // Prints 2, as the object remained on Tuesday
}

输出：

1
2
Exception Out of range
2