将具体实现作为通用返回

Question

我有一个看起来像这样的界面:

public interface IFaker
{
    Faker<T> GetFaker<T>() where T : class;
}

public class DogFaker : IFaker
{
    public Faker<T> GetFaker<T>() where T : class
    {
        return new Faker<Dog>()
            .RuleFor(dog => dog.Name, f => f.Name.FirstName(Name.Gender.Male))
            .RuleFor(dog => dog.IsGoodBoy, f => f.Random.Bool());
    }
}

问题是 - 我不能Faker<Dog>代替Faker<T>例如

Bogus.Faker<_scratchpad.models.Dog>' to 'Bogus.Faker<T>

显然我无法改变new Faker<Dog>,new Faker<T>因为那时我无法访问Dog属性.

我想像这样使用对象:

private Dictionary<string, object> _fakers;

public FakerService()
{
    _fakers = InitialiseFakers();
}

private Dictionary<string, object> InitialiseFakers()
{
   return new Dictionary<string, object>()
    {
        {typeof(Dog).FullName, new DogFaker().GetFaker<Dog>()}
    };
}

public IEnumerable<T> Generate<T>(int count) where T : class
{
    var faker = GetFaker<T>();
    return faker.Generate(count);
}

private Faker<T> GetFaker<T>() where T : class
{
    var faker = (Faker<T>)_fakers[typeof(T).FullName];
    return faker;
}

Answer 1

我建议从以下开始:

public interface IFaker<T> where T : class
{
    Faker<T> GetFaker();
}

public class DogFaker : IFaker<Dog>
{
    public Faker<Dog> GetFaker() 
    {
        return new Faker<Dog>()
        // whatever else you need here
        ;
    }
}

这种方式DogFaker.GetFaker()将其T约束明确定义为Dog- 而不是由调用者定义.