[英]What is the difference between member val and member this in F#?
當我在F#中創建一個包含通用的,可變的.NET Stack的類時,如下例所示,該堆棧忽略了我推送它的任何東西。
open System.Collections.Generic
type Interp(code: int array) =
member val PC = 0 with get, set
member this.stack: Stack<int> = new Stack<int>()
member this.Code = code
let interp = Interp([|1;2;3|])
interp.stack.Push(1)
printfn "%A" interp.stack // prints "seq[]" WAT?!
然而,如果我通過屬性使堆棧可變:
open System.Collections.Generic
type Interp(code: int array) =
member val PC = 0 with get, set
member val stack: Stack<int> = new Stack<int>() with get, set
member this.Code = code
let interp = Interp([|1;2;3|])
interp.stack.Push(1)
printfn "%A" interp.stack // prints "seq[1]"
一切都像我期望的那樣神奇地起作用。
這到底是怎么回事? 我對以前語言(大多數是C#)的不變性的理解會說,即使第一個例子中的堆棧是一個不可變成員,這個不可變性應該只是參考(我不應該重新分配Stack本身)。 我仍然可以將值推送到/從中推送。 我錯過了什么,如果試圖改變該堆棧是錯誤的,為什么不拋出異常或編譯錯誤?
如果您嘗試編譯第一個版本,然后使用例如Reflector將其反編譯為C#,您將看到堆棧成員的定義如下:
public class Interp
{
public Stack<int> stack
{
get { return new Stack<int>(); }
}
// Other members omitted for clarity...
}
如您所見,這也是有效的C#代碼,但顯然不是您想要的。
第二個版本交叉編譯成這樣的東西:
public class Interp
{
internal int[] code;
internal Stack<int> stack@;
public Interp(int[] code) : this()
{
this.code = code;
this.stack@ = new Stack<int>();
}
public Stack<int> stack
{
get { return this.stack@; }
set { this.stack@ = value; }
}
// Other members omitted for clarity...
}
這看起來更像是你想要財產做的事情。
做一個你想要的更慣用的方法是:
open System.Collections.Generic
type Interp(code: int array) =
let stack = Stack<int>()
member val PC = 0 with get, set
member this.Stack = stack
member this.Code = code
如果您不需要在外部公開堆棧,請省略最后一行。
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