通用函数接受类型参数 <T<U> &gt;？

Question

I'm refactoring the following code to use generic to reduce the methods:

class CInt  { public int P1 { get; set; } }
class CDate { public DateTime P1 { get; set; } }
class CStr  { public string P1 { get; set; } }

static IEnumerable<CInt>  Fun(IEnumerable<CInt> p) { .... }
static IEnumerable<CDate> Fun(IEnumerable<CDate> p) { .... }
static IEnumerable<CStr>  Fun(IEnumerable<CStr> p) { .... }

To:

interface IBase<T> { T P1 { get; set; } }
class CInt  : IBase<int> { public int P1 { get; set; } }
class CDate : IBase<DateTime> { public DateTime P1 { get; set; } }
class CStr  : IBase<string> { public string P1 { get; set; } }

static IEnumerable<T> Fun<T, U>(IEnumerable<T> p) where T : IBase<U> { 
    //.... 
    var t = p.First().P1;
    //....
    return null;
}
var cints = new List<CInt> { new CInt() };
var result = Fun1<IBase<int>, int>(cints);

Is it possible to remove the type parameter U in above function since T can decide what U is. Or something like

static IEnumerable<IBase<S>> Fun<IBase<S>>(IEnumerable<IBase<S>> p) { .... }

Or it's not possible unless C# support higher kinded type?

Answer 1

Not sure what you mean by "unless C# support higher order type", but one possible solution is to have IBase<T> extend a non-generic IBase and use that for your constraint on Fun :

interface IBase { }
interface IBase<T> : IBase { T P1 { get; set; } }

static IEnumerable<T> Fun<T>(IEnumerable<T> p) where T : IBase { .... }

Of course, if you use the generic type of IBase<T> in Fun then you have to declare is as a second generic parameter.