2

如果我在非通用构建器类中使用隐式运算符,一切都可以:

 public class ReligionBuilder
{
    private Religion _religion;

    public ReligionBuilder()
    {
        _religion = new Religion(){//some codes}

    }

    public ReligionBuilder AddToRepository()
    {
        Repository<Religion>.Add(_religion);
        return this;
    }
    public Religion Build()
    {
        return _religion;
    }

    public static implicit operator Religion(ReligionBuilder _builder)
    {
        return _builder.Build();

    }
}

我可以使用它:

 Religion religion=new ReligionBuilder().AddToRepository();

但如果这个运算符在泛型类中,则出现问题:

 public abstract class DataTestBuilderBase<T> : IDataTestBuilder<T>
{
    protected T TestData { get; set; }

    public virtual T Build()
    {
        return TestData;
    }

    public abstract IDataTestBuilder<T> AddToRepository();
    public abstract IDataTestBuilder<T> WithDefault();

    public static implicit operator T(DataTestBuilderBase<T> builder)
    {
        return builder.Build();
    }
}


 public class PersonDataTestBuilder : DataTestBuilderBase<Person>
{
    private Person _person;

    public PersonDataTestBuilder()
    {
        //some codes
    }
    public override IDataTestBuilder<Person> AddToRepository()
    {
       //some codes
        return this;
    }
}

用法:

 PersonDataTestBuilder _testBuilder = new PersonDataTestBuilder();
        Person person = _testBuilder.AddToRepository();

错误是:无法将 IDataTestBuilder 转换为 Person

问题是什么?

4

1 回答 1

0

AddToRepository仅返回IDataTestBuilder<Person>(就编译时返回类型而言) - 并且没有从隐式转换为Person. 如果您将抽象方法的返回类型更改为DataTestBuilderBase<T>then 它应该可以工作 - 尽管坦率地说我不想使用那种隐式转换。我通常对提供隐式转换非常谨慎——它们通常会使代码不那么清晰,我相信它们在这里会这样做。

AddToRepository无论如何,您真的应该提供建筑商吗?对于建筑商来说,这感觉像是一种不恰当的行为——我希望:

Person person = new PersonBuilder { /* properties */ }
                     .Build()
                     .AddToRepository();

编辑:只是为了说明我改变返回类型的意思AddToRepository,这里有一个简短但完整的程序来演示。它工作正常。

using System;

public abstract class BuilderBase<T>
{
    public abstract T Build();
    public abstract BuilderBase<T> AddToRepository();

    public static implicit operator T(BuilderBase<T> builder)
    {
        return builder.Build();
    }
}

public class TestBuilder : BuilderBase<string>
{
    public override string Build()
    {
        return "Built by Build()";
    }

    public override BuilderBase<string> AddToRepository()
    {
        return this;
    }
}

class Program
{
    static void Main(string[] args)
    {
        string x = new TestBuilder().AddToRepository();
        Console.WriteLine(x);
    }
}
于 2012-01-16T07:27:23.833 回答