struct vs built-in type

[Rikki Cattermole via Digitalmars-d-learn]

On 17/12/2014 8:57 p.m., Jack Applegame wrote:
> Code:
>
> import std.stdio;
>
> struct Bar {
>      int payload;
>      alias payload this;
> }
> struct Foo {
>      private {
>          Bar m_bar;
>          int m_baz;
>      }
>      @property {
>          Bar bar() { return m_bar; }
>          void bar(Bar v) { m_bar = v; }
>          int baz() { return m_baz; }
>          void baz(int v) { m_baz = v; }
>      }
> }
>
> void main() {
>      Foo foo;
>      foo.bar += 4; // Compiles. Why?
>      foo.baz += 4; // Error: foo.baz() is not an lvalue
> }
>
> foo.baz() is not lvalue. It's true.
> But foo.bar() is lvalue. Why? I expected the same error. This line has
> no effect, but the compiler does not issue even a warning.
>
> DMD64 D Compiler v2.066.1. CentOS 7

In this case, Bar can be treated as a reference type.
Where as a primitive type such as int is not. It is a literal.

Although I swear remembering this was a compiler bug and there was a 
discussion about it.

But just for thought:
struct Foo {
	string text;	
}

void main() {
	Foo foo;
	foo.doit;	
	
	import std.stdio;
	writeln(foo.text);
}


Given definition of doit:
void doit(Foo foo) {
	foo.text = "hi";	
}

It'll output a new line and thats it.
But for this:
void doit(ref Foo foo) {
	foo.text = "hi";	
}

It'll output hi.
In this case its explicitly provided by reference not as a literal.