如果我有下一个类型:
rotateControl: true
我做了:
type Color(r: float, g: float, b:float) =
member this.r = r
member this.g = g
member this.b = b
static member ( * ) (c1:Color, c2:Color) =
Color (c1.r*c2.r, c1.g*c2.g, c1.b*c2.b)
static member Zero = Color(0.0,0.0,0.0)
我应该获得 true ,但是通过脚本进行的F#交互式给我 false 相反,如果我定义为:
let ca = Color(1.,1.,1.)
let cb = Color(1.,1.,1.)
ca = cb
返回 true 我是否在尝试以这种方式比较定义类型的两个值时做错了什么? 我怎样才能获得 true ?
由于
答案 0 :(得分:4)
Color
的OP定义是类。默认情况下,类具有引用相等,就像在C#中一样。这意味着如果它们确实是相同的对象(指向相同的内存地址),它们就是相等的。
只有F#中的功能数据类型具有结构相等。这些包括记录,受歧视的联盟,列表和一些其他类型。
将Color
定义为记录更加惯用:
type Color = { Red : float; Green : float; Blue : float }
此类型具有内置的结构相等性:
> let ca = { Red = 1.; Green = 1.; Blue = 1. };;
val ca : Color = {Red = 1.0;
Green = 1.0;
Blue = 1.0;}
> let cb = { Red = 1.; Green = 1.; Blue = 1. };;
val cb : Color = {Red = 1.0;
Green = 1.0;
Blue = 1.0;}
> ca = cb;;
val it : bool = true
如果要为类型定义乘法和零,也可以这样做:
let (*) x y = {
Red = x.Red * y.Red
Green = x.Green * y.Green
Blue = x.Blue * y.Blue }
let zero = { Red = 0.0; Green = 0.0; Blue = 0.0 }
这使您可以编写,例如:
> let product = ca * cb;;
val product : Color = {Red = 1.0;
Green = 1.0;
Blue = 1.0;}
答案 1 :(得分:3)
F#实现记录和联合的自动成员比较,但不实现类。如果您想拥有它并使用Color(r, g, b)
语法构造值,则可以使用单例联合。您将获得模式匹配作为奖励(请参阅我对(*)
的实现)。
type Color =
| Color of r: float * g: float * b: float
member this.r = let (Color(r, _, _)) = this in r
member this.g = let (Color(_, g, _)) = this in g
member this.b = let (Color(_, _, b)) = this in b
static member (*) (Color(r1, g1, b1), Color(r2, g2, b2)) =
Color(r1 * r2, g1 * g2, b1 * b2)
static member Zero = Color(0., 0., 0.)
答案 2 :(得分:-2)
首先,您应该阅读本页:
http://blogs.msdn.com/b/dsyme/archive/2009/11/08/equality-and-comparison-constraints-in-f-1-9-7.aspx
它很好地说明了F#
中的平等如何运作。
关于您的具体问题,您正在研究参考平等与结构平等之间的区别。您可以添加以下注释
[<CustomEquality; CustomComparison>]
您可以向Equals方法override x.Equals(other)
添加重载以进行成员比较