F# Immutable Class Interop -

how f# immutable types interface c#. i'm starting learn f# , i'd mix in c# code have, want f# classes immutable.

let's we're making vector class in f#. vector.x , vector.y should re-assignable, returning new vector class. in c# take allot of legwork make .withx(float x) clone existing object , return new one. there easy way in f#?

i've been searching time , can't seem find docs on this. great.

and finally, if imported class c# interface like? f# code restrict me doing stupid vector.x = 10?

this similar regardless of whether it's c# or f#.

you "in c# take legwork", cmon, think

vector withx(float x) { return new vector(x, this.y); } 

is it, right?

in both c# , f#, prevent assignment x property, author property 'getter' no 'setter'.

i think you're making of out harder is, or maybe i'm misunderstanding you're asking.

edit

for (i think rare) case of there 20 field , may want change small arbitrary subset of them, found cute hack use f# , c# optional parameters nicely.

f# code:

namespace global  open system.runtime.interopservices  type util =     static member some<'t>(x:'t) = x  type myclass(x:int, y:int, z:string) =     new (toclone:myclass,           [<optional>] ?x,           [<optional>] ?y,           [<optional>] ?z) =              myclass(defaultarg x toclone.x,                      defaultarg y toclone.y,                      defaultarg z toclone.z)     member this.x = x     member this.y = y     member this.z = z 

f# client code:

let = new myclass(3,4,"five") let b = new myclass(a, y=44)  // clone change y 

c# client code:

var m = new myclass(3, 4, "five"); var m2 = new myclass(m, y:util.some(44)); // clone m change y 

that is, optional parameters nice way this, , while c# optional parameters have limitations, can expose f# optional parameters in way works ok c#, suggested above.