为什么茉莉间谍无法通过引用解析功能对象? [英] Why jasmine spy doesn't resolve the function object by reference?

问题描述

我有以下简单服务

app.factory('Shoes', function() {
    function a() {return 12;}
    function b() {return a();}

    return {
      a: a,
      b: b
    }
  })

我想测试在调用方法 b 时是否正在调用方法 a .我的测试如下:

I want to test if the method a is being called when I call method b. My test looks like this:

describe('Testing a Shoes service', function() {
  var service;

  beforeEach(module('plunker'));

  beforeEach(inject(function(Shoes) {
    service = Shoes;
  }))

  it('.b should call .a', function() {
    spyOn(service, 'a');
    service.b();
    expect(service.a).toHaveBeenCalled();
  })

});

但是测试失败.相关的插件在此处.我已经从这些答案中了解了如何解决问题..仍然有一个未解决的问题:

But the tests fail. Relevant plunker is here. I've already know how to resolve the problem in the plunker from those answers. Still there is one unresolved question:

为什么不解析原始功能对象?

我认为系统是这样工作的(假设间谍用附加逻辑来装饰功能)

I thought that the system works like this (assuming spyies decorate the function with additional logic)

当我调用 service.a 时没有间谍时，它被解析为:

When I have no spies when I call service.a it's being resolved as:

A) service.a -> a()

当我创建间谍时，它会装饰功能

and when I create a spy, it decorates the function

B) spy -> service.a => service.a*()

但是 service.a 基本上是原始 a()的引用，因此最后我们应该为已解析的函数对象设置一个间谍:

but the service.a is basically a reference for original a() so we should have a spy set for resolved function object in the end:

A + B => spy -> service.a -> a => a*()

推荐答案

调用 spyOn(service，'a')后， service.a 不再是您在函数中定义的 a -这是一个间谍函数.该间谍函数恰巧调用了 a ，但它不是 a ；它具有不同的身份，并且具有不同的功能.

After you call spyOn(service, 'a'), service.a is no longer the a you defined in the function -- it is a spy function. That spy function happens to call a, but it is not a; it has a different identity and is a different function.

但是，设置 service 的 a 属性不会更改您在 app.factory function a >.您只需更改 service ，以使其 a 属性不再引用原始的 a .相比之下，您的 b 函数永远不会将其引用更改为原始的 a .它直接从最初声明了 a 和 b 的本地 app.factory 范围获取 a . service 用间谍替换其原始 a 的事实不会影响 b 对 a()的调用，因为它没有引用 service.a .

However, setting the a property of service does not change the function a you declared inside app.factory. You've simply changed service so that its a property no longer refers to your original a. By contrast, your b function never changes its reference to the original a. It gets a straight from the local app.factory scope in which a and b were originally declared. The fact that service replaces its original a with a spy does not affect b's call to a(), because it does not refer to service.a.

这篇关于为什么茉莉间谍无法通过引用解析功能对象?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！