Does the IS operator unbox value type or not?

Question

I can't find an answer to the following question:

object o = 10; // Box
int i = (int)o; // Unbox

it's clear, but the following isn't clear

bool isInt = o is int; // Is the unbox here or not?

Answer 1

No, that's not unboxing - it's just checking whether the type is correct. Don't forget that there really is an object involved, with a type. Checking that type is basically the same operation regardless of whether the value is a boxed value type value or not. (There may be some optimizations feasible for value types or any sealed types, as there's no inheritance to consider, but fundamentally it's still checking the "type" part of an object header.)

One way to check that is to compile the code and look at the IL using ILASM:

// object o = 10
IL_0000:  ldc.i4.s   10
IL_0002:  box        [mscorlib]System.Int32
IL_0007:  stloc.0

// int i = (int) o;
IL_0008:  ldloc.0
IL_0009:  unbox.any  [mscorlib]System.Int32
IL_000e:  stloc.1

 // bool isInt = o is int
IL_000f:  ldloc.0
IL_0010:  isinst     [mscorlib]System.Int32

So it uses isinst - no unboxing is necessary.