Re: 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.