如何生成任意类型的存根对象不使用AutoFixture在编译时已知 [英] How to generate a stub object of an arbitrary Type not known at compile time using AutoFixture

问题描述

我能得到这样的构造函数参数的类型：

I can get type of constructor parameter like this:

Type type = paramInfo.ParameterType;

现在我想从这种类型的创建存根对象。是有可能吗？我试着用autofixture：

Now I want to create stub object from this type. Is is possible? I tried with autofixture:

public TObject Stub<TObject>()
{
   Fixture fixture = new Fixture();   
   return fixture.Create<TObject>();
}

..但它不工作：

.. but it doesn't work:

Type type = parameterInfo.ParameterType;   
var obj = Stub<type>();//Compile error! ("cannot resolve symbol type")

你能帮我吗？

Could you help me out?

推荐答案

AutoFixture的确实的有一个非通用API来创建对象，的尽管那种隐藏的（设计）：

AutoFixture does have a non-generic API to create objects, albeit kind of hidden (by design):

var fixture = new Fixture();
var obj = new SpecimenContext(fixture).Resolve(type);

作为的由@meilke链接博客帖子指出，如果你发现自己需要这通常，你可以在一个扩展方法封装它：

As the blog post linked by @meilke points out, if you find yourself needing this often, you can encapsulate it in an extension method:

public object Create(this ISpecimenBuilder builder, Type type)
{
    return new SpecimenContext(builder).Resolve(type);
}

允许你简单地做：

which allows you to simply do:

var obj = fixture.Create(type);

这篇关于如何生成任意类型的存根对象不使用AutoFixture在编译时已知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！