[英]How to add members to a class dynamically using options as coonstructor args in python?
[英]Define several members dynamically in class/interface using F#
我正在使用F#进行小型程序开发,仅用于个人培训,但花了点时间,我无法解决。
我正在描述一些接口:
type IСalculations =
abstract member Add : int * int -> int
abstract member Subtract : int * int -> int
abstract member Multiply : int * int -> int
abstract member Divide : int * int -> int
如您所见,除名称外,成员签名是相同的。
我能否在下一个中使用F#(现在将是伪代码 ):
let names = [ "Add", "Subtract", "Multiply", "Divide" ];
let ICalculations = new interface;
foreach ( name in names ) {
ICalculations[ name ] : int * int -> int
}
目的不是要为每个成员重复签名int * int -> int
可能吗?
接口声明后不能定义接口方法类型。 但是您可以定义例如Dictionary,其中包含您类型的函数:
open System.Collections.Generic
type Ops = Add | Subtract | Multiply | Divide
let pseudoInterface = Dictionary<Ops, int * int -> int>()
// Then somewhere in your program you could define this "methods"
pseudoInterface.[Add] <- fun (i, j) -> i + j
pseudoInterface.[Subtract] <- fun (i, j) -> i - j // etc...
或者,为了简洁起见,您可以为函数类型定义类型别名:
type Op = int * int -> int
type IСalculations =
abstract member Add : Op
abstract member Subtract : Op
abstract member Multiply : Op
abstract member Divide : Op
声明接口的唯一语法是:
// Interface declaration:
[ attributes ]
type interface-name =
[ interface ] [ inherit base-interface-name ...]
abstract member1 : [ argument-types1 -> ] return-type1
abstract member2 : [ argument-types2 -> ] return-type2
...
[ end ]
在伪代码的第二行:
let ICalculations = new interface;
您希望使用let绑定或等效绑定。 不幸的是,让绑定仅将标识符与值或函数相关联,而不与类型或接口相关联。 因此,恐怕没有办法。 F#之外的其他功能语言(例如Idris)也可以做到。 如果您只被重复int * int-> int的冗长困扰,则可以定义类型别名,如下所示:
module T =
type Op = int * int -> int
type IСalculations =
abstract member Add : T.Op
abstract member Subtract : T.Op
abstract member Multiply : T.Op
abstract member Divide : T.Op
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.