abh*_*hek 10 f# restriction f#-3.0
我试图将我的语法定义为一个有区别的联盟.它有两种可能:int与datetime和mathemetical运营商Add和Mul.
Add适用于int和datetime(如加在INT天)
Mul只在工作int,而不是datetime
语法可以递归
我的语法看起来像
type MyExpression =
|Integer of int
|Date of datetime
|Add of MyExpression * MyExpression
|Mul of MyExpression * MyExpression
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我编写了一个解析器(fparsec),它可以解析我的语法中的文本,但我不知道如何处理Mul可递归的条件,但仅限于Integer.
是否可以选择在我的MyExpression类型上定义此限制,还是必须在我的解析输入中处理此问题?
阿斯蒂建议的设计也是我的首选.根据您的要求,这可能就是您所需要的.
但是,它确实可以编译像这样的表达式
Add(Val(System.Console.Out), Val(System.Console.Error))
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这可能不是你想要的.
或者,您可以像这样建模表达式:
open System
type IntExpression =
| Integer of int
| Mul of IntExpression * IntExpression
| Add of IntExpression * IntExpression
type DateTimeExpression =
| Date of DateTime
| Add of DateTimeExpression * DateTimeExpression
type MyExpression =
| IntExpression of IntExpression
| DateTimeExpression of DateTimeExpression
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这显然是一个更详细的类型定义,但它确实体现了一个表达式可以包含整数或DateTime值的叶节点而不包含其他值的规则 - 如果这是您要强制执行的规则.
我并不是说这更好; 我只提供替代品.
用法:
> IntExpression(Mul(IntExpression.Add(Integer(1), Integer(2)),Integer 3));;
val it : MyExpression =
IntExpression (Mul (Add (Integer 1,Integer 2),Integer 3))
> DateTimeExpression(Add(Date(DateTime.MinValue),Date(DateTime.MinValue)));;
val it : MyExpression =
DateTimeExpression
(Add
(Date 01.01.0001 00:00:00 {Date = 01.01.0001 00:00:00;
Day = 1;
DayOfWeek = Monday;
DayOfYear = 1;
Hour = 0;
Kind = Unspecified;
Millisecond = 0;
Minute = 0;
Month = 1;
Second = 0;
Ticks = 0L;
TimeOfDay = 00:00:00;
Year = 1;},
Date 01.01.0001 00:00:00 {Date = 01.01.0001 00:00:00;
Day = 1;
DayOfWeek = Monday;
DayOfYear = 1;
Hour = 0;
Kind = Unspecified;
Millisecond = 0;
Minute = 0;
Month = 1;
Second = 0;
Ticks = 0L;
TimeOfDay = 00:00:00;
Year = 1;}))
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如果您有基于类型的约束,那么通用方法可能更容易:
type MyExpression<'t> =
|Val of 't
|Mul of MyExpression<int> * MyExpression<int>
|Add of MyExpression<'t> * MyExpression<'t>
let Integer (x:int) = Val(x)
let Date (x:DateTime) = Val(x)
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用法:
Mul(Integer(1), Integer(2)) //compiles
Mul(Date(DateTime.Now), Date(DateTime.Now)) //error
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