I need to implement a static extension method supporting member constraints on some basic primitive types like integers, floats, etc. Here's my code for signed integers:
module MyOperators =
let inline foo (x : ^T) = (^T : (static member Foo : ^T -> int) (x))
type System.Int32 with
static member Foo(x : Int32) = 7 // static extension
Test code:
open MyOperators
let x = foo 5 // x should be 7
But compiler complains with error:
The type …
我理解用于表示通用参数的刻度,如:
Seq.append : seq<'T> -> seq<'T> -> seq<'T>
但是插入符号表示什么,如:
Seq.average : seq<^T> -> ^T
以下(简化)代码段取自我正在实现的应用程序,该应用程序始终使用静态解析的类型参数.
type A< ^B when ^B : (static member MyMember : Unit -> Unit)> = {
Field : unit
}
type TestA = {
AField : A< BTy >
}
and BTy = {
BField : Unit
} with
static member MyMember () = ()
当我定义字段AField(AField : A< BTy >)的类型时,IntelliSense给出了以下错误:类型'BTy'不支持任何名为'MyMember'的运算符.
编辑.单独声明它们是有效的,但如果我有一个共同引用,我不能声明第三种类型放在顶部,其中包含两种类型的公共信息.我该怎么做才能避免这个问题？无论如何,如果我在下面定义let pluto = ("" :> obj) :?> A< BTy >它的定义,我想是因为这两种类型都是从let绑定中可见的.
我一直在寻找一种方法来为F#方法添加一些鸭子类型.
SomeMethod(model:'a) =
let someField = model.Test("")
进来的参数有Test方法.我看过这样的符号:
member inline public x.Testing< ^a when ^a : (member public Test : String-> String)>(model:^a) =
let something = model.Test("")
ignore
对我来说,通用约束可以用于方法级别而不是类/接口级别.问题是由于类型问题我无法编译它.这让我相信没有办法在方法级别指定约束.这是coorect？
module FSharp=
let Point2d (x,y)= Point2d(x,y)
let Point3d (x,y,z)= Point3d(x,y,z)
type NXOpen.Point3d with
static member ( * ) (p:Point3d,t:float)= Point3d(p.X*t,p.Y*t,p.Z*t)
static member ( * ) (t:float,p:Point3d)= Point3d(p.X*t,p.Y*t,p.Z*t)
static member (+) (p:Point3d,t:float)= Point3d(p.X+t,p.Y+t,p.Z+t)
static member (+) (t:float,p:Point3d)= Point3d(p.X+t,p.Y+t,p.Z+t)
static member (+) (p:Point3d,t:Point3d)= Point3d(p.X+t.X,p.Y+t.Y,p.Z+t.Z)
let a=Point3d (1.,2.,3.)
let b=1.0
let c=a * b//error
错误15:类型'float'与
'Point3d' 类型不匹配E:\ Work\extension-RW\VS\extension\NXOpen.Extension.FSharp\Module1.fs 18 13 NXOpen.Extension.FSharp
我想扩展Point3d方法,一些新的运算符.但它并没有过去.
我想我知道目标从句和号角从句的含义，但是我对人们为什么命名一个不带正向目标的字面量的歧义感到困惑。
这里的目标是什么？
我是F#的新手,来自C++背景.我正在尝试编写一个简单的向量类,它可以是泛型类型(int,float等),但我遇到了默认构造函数的问题.我想将值初始化为零,但为了做到这一点,我需要以某种方式将一个具体的零转换为泛型类型,但我不知道如何做到这一点.
也许一些代码可能有所帮助 这是我到目前为止所拥有的:
type Vector3D<'T> (x :'T, y: 'T, z: 'T) =
member this.x = x
member this.y = y
member this.z = z
new() = Vector3D<'T>(0,0,0) // what goes here?
我在突出显示的行上尝试了很多东西,但似乎无法让编译器感到高兴.例如,我试过,Vector3D('T 0, 'T 0, 'T 0)我认为应该将int零'T归零,但这不起作用.
我错过了一些基本的东西,还是仅仅是获得正确语法的情况？
在以下代码中,请注意get_Zero的类型约束:
type Wrapper<'t> = { Data : 't[] }
let compute<'t
when 't : (static member get_Zero : unit -> 't)
and 't : (static member (~-) : 't -> 't)
and 't : (static member (+) : 't * 't -> 't)>
(wrapper : Wrapper<'t>) =
wrapper.Data
|> Seq.mapi (fun i value -> (i, value))
|> Seq.sumBy (fun (i, value) ->
if i % 2 = 0 then value
else -value)
即使我已经有一个显式类型约束,我仍然在调用Seq.sumBy时遇到以下编译器错误:
类型参数缺少约束'when ^ t :(静态成员get_Zero: - > ^ …
为什么不允许这样做？
type Foo() =
static member Bar() = ()
let inline bar<^a>() = //ERROR: unexpected infix operator in pattern
(^a : (static member Bar : unit -> unit)())
//Hypothetical usage: let _ = bar<Foo>()
......但是这样可以吗？
type Foo() =
static member Bar() = new Foo()
let inline bar() : ^a =
(^a : (static member Bar : unit -> ^a)())
let x : Foo = bar()
具有静态解析类型参数的函数是否需要返回已解析类型的实例？
在F#中,如何编写泛型数学步骤函数？
(Oliver)Heaviside阶跃函数是函数,如果x为负,则返回零,否则返回一个.
以下是我到目前为止的尝试摘要:
// attempt 1:
let inline stepFct1< ^T when ^T : (static member op_GreaterThan: ^T * float -> bool)
> (x:^T) : ^T =
//if (^T : (static member op_GreaterThan) (x 0.0) ) then x //ouch fails also
if (>) x 0.0 then x
else 0.0
编译器说:错误FS0001:类型参数缺少约束'当^ T:比较'
// attempt 2:
let inline stepFct2<^T when ^T : (static member (>): ^T * ^T -> bool) > (x:^T) : ^T =
match x with
| x when x > …为了好玩,我一直在玩F#中的类型类,使用这里显示的想法.我创建了一个Next类型类来表示具有"后继"的值,例如下一个1 = 2,下一个今天=明天等:
type Next = Next with
static member (++) (Next, x:int) = x + 1
static member (++) (Next, x:DateTime) = x.AddDays 1.0
let inline next x = Next ++ x
let v1 = next 1 // v1 = 2
let v2 = next DateTime.Now // v2 = Now + 1 day
现在我想在泛型类中使用"nextable"类型:
type UsesNextable<'T>(nextable: 'T) = // Compile error 1
member inline this.Next =
let v = next nextable // Compile error 2
v.ToString()
但是我收到以下错误: …