使用let语句中提供的midje不会存根方法 [英] Using midje provided in a let clause doesn't stub methods

问题描述

当我使用一个外部let子句来构造一些定义和调用的测试时，存根不会按照我期望的方式工作。例如：

此测试失败

  
（let [x（meth1 123）] 
x => 246 
（提供
（meth2 123）=> 246）））
  
 
 
 
使用此代码
  meth2 [x] 
（prnmeth2x）
（* 3 x））
 
（defn meth1 [x] ）] 
y））
  
我不应该使用 let 有midje的声明？我不明白如何让这些通过，而不删除 let 。
解决方案
首先，如果 meth2 在（fn [] x）返回一个函数，因此事实（fn [] x）=> 
 
 
 
其次，我认为提供用于存根函数调用只有在执行事实的左/右侧时执行。在你的情况下，（fn [] x）（和 x 已经在那一点评估）  246 （这是常量）。  meth1 永远不会在您的事实上下文中被调用。
 
 
 
的两件事。或者你让 let 部分在你的事实的左边：
  ... 
（let [x（meth1 123）] x）=> 246 
 ... 
  
或者让 x 在测试事实时评估的函数：
  ... 
 ＃（meth1123）]（x））= 246 
 ... 
  
我不认为有办法真的 
 提供 >（let [x（meth1 123）] 
x => 369 
（let [...] 
x => 246 
（提供
 ）=> 246））））
  
 A  let 包装事实似乎在触摸第一个事实之前执行。
 
When I make a test using an outer let clause to structure some definitions and calls, the stubs don't work the way I'd expect. For example: 

This test fails 
(fact "blah"
  (let [x (meth1 123)]
    x => 246
    (provided
      (meth2 123) => 246)))
With this code 
(defn meth2 [x]
  (prn "meth2" x)
  (* 3 x))

(defn meth1 [x]
  (let [y (meth2 x)]
    y))
Am I not supposed to use let statements with midje? I can't understand how to get these to pass without removing the let.
解决方案
First of all, your test would even fail if meth2 was stubbed correctly since (fn [] x) returns a function, thus the fact (fn [] x) => 246 will never hold.

Secondly, I think that provided is used to stub function calls only when executing the left/right-hand-side of facts. In your case that's (fn [] x) (and x is already evaluated at that point), as well as 246 (which is constant). meth1 never gets called in the context of your fact, only beforehand.

To change that, you could do one of two things. Either you make the let part of the left-hand-side of your fact:
...
(let [x (meth1 123)] x) => 246
...
Or you make x a function that is evaluated when the fact is tested:
...
(let [x #(meth1 123)] (x)) => 246
...
I don't think there is a way to really see provided in action, i.e. like this:
(let [x (meth1 123)]
   x => 369
  (let [...]
    x => 246
    (provided 
      (meth2 123) => 246))))
A let wrapping facts seems to be executed before the first fact is touched.

                        
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