后代可见的公共类成员

发布于 2024-07-23 05:07:59 字数 957 浏览 6 评论 0原文

作为另一个类的公共类成员，我从构建器模式中获得了很大的吸引力：

public class Part
{
    public class Builder
    {
        public string Name { get; set; }
        public int Type { get; set; }

        public Part Build()
        {
            return new Part(Name, Type);
        }
    }

    protected Part(string name, int type)
    {
        ...
    }
}

注意受保护的构造函数 - 我喜欢如何使用构建器来获取零件。呼吁

Part p = new Part.Builder() { Name = "one", Type = 1 }.Build();

工作很棒。我想做的是使用这个构建器根据类型（例如）提供一种特殊类型的零件：

public class SpecialPart : Part
{
    protected SpecialPart(string name, int type) : base(name, type) { }
}

并对构建器进行轻微更改：

public Part Build()
{
    if (Type == _some_number_)
        return new SpecialPart(Name, Type);
    return new Part(Name, Type);
}

但这不起作用 - Part.Builder 无法看到 SpecialPart 的受保护的构造函数。如何让 Builder 与 Part 的后代一起工作并获得相同的必须具有构建器语义？

原文

I have been getting a lot of traction from a builder pattern as a public class member of another class:

public class Part
{
    public class Builder
    {
        public string Name { get; set; }
        public int Type { get; set; }

        public Part Build()
        {
            return new Part(Name, Type);
        }
    }

    protected Part(string name, int type)
    {
        ...
    }
}

Note protected constructor - I like how I HAVE to use the builder to get a Part. Calls to

Part p = new Part.Builder() { Name = "one", Type = 1 }.Build();

work great. What I would like to do is use this builder to serve up a special kind of part based on the Type (for example):

public class SpecialPart : Part
{
    protected SpecialPart(string name, int type) : base(name, type) { }
}

And a slight change to the builder:

public Part Build()
{
    if (Type == _some_number_)
        return new SpecialPart(Name, Type);
    return new Part(Name, Type);
}

But this doesn't work - Part.Builder can't see SpecialPart's protected constructor. How can I get Builder to work with descendents of Part and get the same must-have-a-builder semantics?

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