Given “where T : new()”, does “new T()” use Activator.CreateInstance internally?

Question

If I have a type parameter constraint new():

void Foo() where T : new()
{
    var t = new T();
}

Is it true that

Answer 1

Yes. It does for reference types.

Using ILSpy on the following release-compiled code:

    public static void DoWork() where T: new()
    {
        T t = new T();
        Console.WriteLine(t.ToString());
    }

Yielded

.method public hidebysig 
    instance void DoWork<.ctor T> () cil managed 
{
    // Method begins at RVA 0x2064
    // Code size 52 (0x34)
    .maxstack 2
    .locals init (
        [0] !!T t,
        [1] !!T CS$0$0000,
        [2] !!T CS$0$0001
    )

    IL_0000: ldloca.s CS$0$0000
    IL_0002: initobj !!T
    IL_0008: ldloc.1
    IL_0009: box !!T
    IL_000e: brfalse.s IL_001b

    IL_0010: ldloca.s CS$0$0001
    IL_0012: initobj !!T
    IL_0018: ldloc.2
    IL_0019: br.s IL_0020

    IL_001b: call !!0 [mscorlib]System.Activator::CreateInstance()

    IL_0020: stloc.0
    IL_0021: ldloca.s t
    IL_0023: constrained. !!T
    IL_0029: callvirt instance string [mscorlib]System.Object::ToString()
    IL_002e: call void [mscorlib]System.Console::WriteLine(string)
    IL_0033: ret
} // end of method Program::DoWork

Or in C#:

public void DoWork() where T : new()
{
    T t = (default(T) == null) ? Activator.CreateInstance() : default(T);
    Console.WriteLine(t.ToString());
}

JIT will create different compiled instructions for each different value type parameter passed in, but will use the same instructions for reference types -- hence the Activator.CreateInstance()