我想在不调用fun2的情况下对fun1进行单元测试. let fun2() = // Some complex function with lots of dependencies. 1let 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
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?
这取决于您的用法,但您可以这样做:let fun2() = // Some complex function with lots of dependencies. 1 let createFun1 fun2 = fun () -> fun2() * 2 let fun1 = createFun1 fun2
这对于单元测试也很有用,因为你可以通过简单地为fun2传递一个简单的函数来测试fun1.