Generic method where T is type1 or type2

Question

Is there a way to declare a generic function that the generic type is of type1 or type2?

example:

public void Foo<T>(T number) { } 

Can I constraint T to be int or long

Answer 1

For ReferenceType objects you can do

public void DoIt<T>(T someParameter) where T : IMyType {  } 

...

public interface IMyType { }  public class Type1 : IMyType { }  public class Type2 : IMyType { } 

For your case using long as parameter will constrain usage to longs and ints anyway.

public void DoIt(long someParameter) {  } 

to constrain to any value types (like: int, double, short, decimal) you can use:

public void DoIt<T>(T someParameter) where T : struct {  } 

for more information you can check official documentation here

Answer 2

Although you could use a generic constraint to limit the type of each generic argument T, unfortunately there is none that would allow you to enforce at compile time whether T is type1 or type2.

Nor is there any way to enforce at compile time that your generic argument can only be of any primitive type (int, long, double, ...).