Why GetType returns System.Int32 instead of Nullable<Int32>? [duplicate]

Question

Why is the output of this snippet System.Int32 instead of Nullable<Int32>?

int? x = 5; Console.WriteLine(x.GetType()); 

Answer 1

GetType() is a method of object.
To call it, the Nullable<T> struct must be boxed.

You can see this in the IL code:

//int? x = 5; IL_0000:  ldloca.s    00  IL_0002:  ldc.i4.5     IL_0003:  call        System.Nullable<System.Int32>..ctor  //Console.WriteLine(x.GetType()); IL_0008:  ldloc.0      IL_0009:  box         System.Nullable<System.Int32> IL_000E:  callvirt    System.Object.GetType IL_0013:  call        System.Console.WriteLine 

Nullable types are treated specially by CLR; it is impossible to have a boxed instance of a nullable type.
Instead, boxing a nullable type will result in a null reference (if HasValue is false), or the boxed value (if there is a value).

Therefore, the box System.Nullable<System.Int32> instruction results in a boxed Int32, not a boxed Nullable<Int32>.

Therefore, it is impossible for GetType() to ever return Nullable<T>.

To see this more clearly, look at the following code:

static void Main() {     int? x = 5;     PrintType(x);    } static void PrintType<T>(T val) {     Console.WriteLine("Compile-time type: " + typeof(T));     Console.WriteLine("Run-time type: " + val.GetType()); } 

This prints

Compile-time type: System.Nullable`1[System.Int32]
Run-time type: System.Int32

Answer 2

GetType() isn't virtual, and is thus defined only on object. As such, to make the call, the Nullable<Int32> must first be boxed. Nullables have special boxing rules, though, so only the Int32 value is boxed, and that's the type reported.