C# reflected type as type parameter

Question

I have a Type, which I parsed from it's fully qualified name:

Type paramType = Type.GetType(param.TypeName);

How can I use paramType to create an instance of an object with paramType as a type parameter?

[Activator.CreateInstance Method @ learn.microsoft.com](https://learn.microsoft.com/dotnet/api/system.activator.createinstance) — , Jul 09 '20 at 11:26
@AlaaMohammed obviously because it's only helps when compiler can determine type at COMPILE TIME — Selvin, Jul 09 '20 at 11:50

score 1 · Accepted Answer · answered Jul 09 '20 at 11:36

1

It's simple:

void Main()
{
    Type typeParam = typeof(int);
    Type typeGeneric = typeof(Foo<>).MakeGenericType(typeParam);
    
    object foo = Activator.CreateInstance(typeGeneric);
    //foo is a Foo<int>
}

public class Foo<T>
{
    
}

answered Jul 09 '20 at 11:36

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C# reflected type as type parameter

1 Answers1