let fun2() =
    // Some complex function with lots of dependencies.
    1

let fun1() =
    fun2() * 2

打破两个函数之间依赖关系的最佳方法是什么？

我尝试了几种不同的方法,但它们只是增加了混乱.

将fun2传递给fun1

let fun1(fun2) =
    fun2() * 2

转换为类并覆盖

type FunClass() =
    abstract member fun2 : unit -> int
    default x.fun2() = 1

    member x.fun1() =
        x.fun2() * 2

type FunClassMock() =
    override member x.fun2() = 1

使用状态模式

type Fun1Class(fun2Class) =

    member x.fun1() =
       fun2Class.fun2() * 2

使用变量

let fun2Imp() =
    1

let mutable fun2 = fun2Imp

let fun1() =
    fun2() * 2

有更干净的方式吗？