OOAD 家具和测试设计 [关闭]答案

【问题标题】：OOAD Design for furniture and testing [closed]OOAD 家具和测试设计 [关闭]
【发布时间】：2014-05-27 10:46:36
【问题描述】：

这个问题是在一次采访中被问到的。

家具可以由一种或多种材料制成。有各种家具类型，如椅子、桌子、沙发等。

家具测试取决于材料和家具类型。很少有测试是 choakable、防火、WithStands100KgWeight 等

在 OOAD 中设计。设计应该遵循Open Close原则，并且应该是乐观的，未来如果添加新家具，应该不需要更多的代码更改。

最初建议使用

class Furniture{
List<Material> materials;
boolean supportsTest(Test test);
}

class Chair extends Furniture{
boolean supportsTest(Test test){
// check material and based on type return true/false
}

面试官说家具就是家具，不应该说支持与否。任何替代解决方案？谢谢。

【问题讨论】：

它被否决的任何原因？

标签： oop ooad

【解决方案1】：

根据你的描述和面试官的评论，我想到的是：

// An interface so that it's separate from Furniture class
interface FurnitureTest {
   bool IsSupportedBy(Furniture furniture);
}

// Sample test, can add as many as you like
class FireResistable : FurnitureTest {
   public bool IsSupportedBy(Furniture furniture) {
       return furniture.Type.Name == ! "Mirror" && /* just to use the Type property */
              !furniture.HasMaterial(x => x == "Wood");
   }
}

///
/// Other class definitions.
///

// Open to extension
class Chair : Furniture { /* additional properties */ }
class Table : Furniture { /* additional properties */ }

// Closed to modification
class Furniture {
   public FurnitureType FurnitureType { get; private set; }
   private IList<Material> materials;

   public Furniture(FurnitureType type, IList<Material> materials) {
      this.materials = materials;
   }

   public bool HasMaterial(string materialName) {
      return materials.Any(x => x.Name == materialName);
   }
}

class FurnitureType {
   public string Name { get; set; }
   // ...other properties
}

class Material {
   public string Name { get; set; }
   // other properties..
}

【讨论】：