10

今天我确实在 D 邮件列表中询问是否可以以类似于 Ada 的wiki页面中的示例的方式定义和使用自定义数据类型：

type Day_type   is range    1 ..   31;
type Month_type is range    1 ..   12;
type Year_type  is range 1800 .. 2100;
type Hours is mod 24;
type Weekday is (Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday); 


type Date is
   record
     Day   : Day_type;
     Month : Month_type;
     Year  : Year_type;
   end record;

subtype Working_Hours is Hours range 0 .. 12;
subtype Working_Day is Weekday range Monday .. Friday;
Work_Load: constant array(Working_Day) of Working_Hours 
   := (Friday => 6, Monday => 4, others => 10);

回复显示如下：

import std.typecons;
import std.exception;

struct Limited(T, T lower, T upper)
{
    T _t;
    mixin Proxy!_t; //Limited acts as T (almost)
    invariant()
    {
        enforce(_t >= lower && _t <= upper);
    }
    this(T t)
    {
        _t = t;
    }
}

auto limited(T, T lower, T upper)(T init = T.init)
{
    return Limited!(T, lower, upper)(init);
}

unittest
{
    enum l = [-4,9];
    auto a = limited!(int, l[0], l[1])();
    foreach(i; l[0] .. l[1]+1)
    {
        a = i;
    }

    assertThrown({a = -5;}());
    assertThrown({a = 10;}());
}

这表明这是可能的，但可能错过了艾达的优雅。

现在，在最近阅读了 Nimrod 之后，我想知道它如何处理类似的任务以确保相同的 Ada 的类型安全？

4

2 回答 2

7

Nimrod 相当直接地支持这些：

type
  Day = range[1..31]
  Month = range[1..12]

  WeekDay = enum
    Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday

  WorkingDays = range[Monday..Friday]
  WorkingHours = range[0..12]

  WorkSchedule = array[WorkingDays, WorkingHours]

错误在编译时强制执行：

var x: Day
x = 40 # conversion from int literal(40) to Day is invalid

.. 或在运行时

var x: Day
var y = unknownInt() # let's say it returns 100

x = y # unhandled exception: value 100 out of range [EOutOfRange]

此外，如果需要更强的类型安全性，可以使用不同的类型。

于 2013-08-30T13:49:49.523 回答
0

此外，实例化成为一个更简单的参数

import std.traits;

/*! Instantiator for \c Limited. */
template limited(alias min, alias max, bool Exceptional = true)
    if (!is(CommonType!(typeof(min), typeof(max)) == void)) {
    auto limited(CommonType!(typeof(min), typeof(max)) value) {
        return Limited!(typeof(value), min, max)(value);
    }
}

请参阅绑定模板的实例化器函数无法编译

于 2013-09-13T20:30:25.363 回答