可以从表达式参数推断其类型参数的 C# 通用方法

Question

I am attempting to create a method, that can accept a type of an object as a type parameter, and a reference to one of the object's properties, along with its type. Something like so:

service.DoWork<DateTime>(dt => dt.Ticks) .

I looked at how other libraries were doing such things, and ended up with the following definition:

interface IService
{
    void DoWork<TObject, TProperty>(Expression<Func<TObject, TProperty>> propertySelector);
}

Ideally, a user can pass in a type of an object, and then "select" each of its properties with an expression. My method can then know the expression selected , along with its type and the type of the object .

  service.DoWork<DateTime>(dt => dt.Ticks);
  service.DoWork<DateTime>(dt => dt.TimeOfDay);
  service.DoWork<DateTime>(dt => dt.Second);

However, calling the method this way results in the following error:

Using the generic method 'IService.DoWork<TObject, TProperty>(Expression<Func<TObject, TProperty>>)' requires 2 type arguments

If I specfiy the type of the property, then everything works fine:

  service.DoWork<DateTime, long>(dt => dt.Ticks);
  service.DoWork<DateTime, TimeSpan>(dt => dt.TimeOfDay);
  service.DoWork<DateTime, int>(dt => dt.Second);

BUT , this seems really redundant. The compiler can extract the type from the expression argument that I have provided, and as far as I know a lot of other libraries use such type inferring to provide clean interfaces (Linq, MOQ, FluentAssertions to name a few)

I imagine i need to somehow change my method definition, but I can't seem to find a correct way to do it.

PS: Here is the complete code snippet.

using System;
using System.Linq.Expressions;

namespace GenericTypeInterface
{
    class Program
    {
        static void Main(string[] args)
        {
            IService service = null;

            service.DoWork<DateTime>(dt => dt.Ticks);
            service.DoWork<DateTime>(dt => dt.TimeOfDay);
            service.DoWork<DateTime>(dt => dt.Second);
        }
    }

    interface IService
    {
        void DoWork<TObject, TProperty>(Expression<Func<TObject, TProperty>> propertySelector);
    }
}

Answer 1

您可以通过明确写入 lambda 参数的类型来让它“推断”两个泛型参数：

service.DoWork((DateTime dt) => dt.Ticks);