C# struct new StructType() vs default(StructType)

Question

Say I have a struct

public struct Foo
{
    ...
}

Is there any difference between

Foo foo = new Foo();

and <

Answer 1

For value-types, the options are, practically speaking, equivalent.

However, I was intrigued by Jon Skeet's empirical research into which 'instructions' result in the invocation of a struct's parameterless default constructor when it is specified in CIL (you can't do it in C# because it doesn't let you). Amongst other things, he'd tried out default(T) and new T() where T is a type parameter. They appeared equivalent; neither of them appeared to call the constructor.

But the one case (it appears) he hadn't tried was default(Foo) where Foo is an actual struct type.

So I took his code for the 'hacked' struct and tried that out for myself.

---

It turns out that default(Foo) doesn't call the constructor, whereas new Foo() in fact does.

Using a struct type Oddity that specifies a parameterless constructor:

With optimizations turned off, the method:

private void CallDefault()
{
    Oddity a = default(Oddity);
}

produces the CIL (without nops, rets etc.):

L_0001: ldloca.s a
L_0003: initobj [Oddity]Oddity

whereas the method:

private void CallNew()
{
    Oddity b = new Oddity();
}

produces:

L_0001: ldloca.s b
L_0003: call instance void [Oddity]Oddity::.ctor()

With optimizations turned on, the compiler appears to optimize away pretty much all of the CallDefault method into a no-op, but keeps the call to the constructor in CallNew (for potential side-effects?).