小编Jav*_*rSA的帖子

我正在使用我的Ubuntu 14.04 LTS进行构建但是我得到以下内容:

Started by user anonymous
Building in workspace /var/lib/jenkins/workspace/videovixx
 > /usr/bin/git rev-parse --is-inside-work-tree # timeout=10
Fetching changes from the remote Git repository
 > /usr/bin/git config remote.origin.url https://bitbucket.org/mdennis10/videovixx.git #     timeout=10
Fetching upstream changes from https://bitbucket.org/mdennis10/videovixx.git
 > /usr/bin/git --version # timeout=10
using .gitcredentials to set credentials
 > /usr/bin/git config --local credential.helper store --    file=/tmp/git6236060328558794078.credentials # timeout=10
 > /usr/bin/git fetch --tags --progress https://bitbucket.org/mdennis10/videovixx.git   +refs/heads/*:refs/remotes/origin/*
 > /usr/bin/git config --local --remove-section credential # timeout=10
 > /usr/bin/git rev-parse refs/remotes/origin/master^{commit} # timeout=10
 > /usr/bin/git rev-parse …

如果我们想在类型上测试扩展函数,我们可以创建这种类型的实例,调用函数并检查返回的值.但是如何测试类中定义的扩展函数呢？

abstract class AbstractClass<T> {
    fun doStuff(): T = "Hello".foo()

    abstract fun String.foo(): T
}

class SubClass1: AbstractClass<Int>() {
    override fun String.foo(): Int = 1
}

class SubClass2: AbstractClass<Boolean>() {
    override fun String.foo(): Boolean = true
}

我们如何测试方法的逻辑foo()中类SubClass1和SubClass2？它甚至可能吗？

我知道我可以改变设计来测试它.我有两种可能性:

1. 不要使用扩展功能.¯\ _(ツ)_ /¯

   abstract class AbstractClass<T> {
       fun doStuff(): T = foo("Hello")
   
       abstract fun foo(string: String): T
   }
   
   class SubClass1: AbstractClass<Int>() {
       override fun foo(string: String): Int = 1
   }
   

   Run Code Online (Sandbox Code Playgroud)

   然后我们可以创建一个对象SubClass1,调用foo()并检查返回的值. …