references and function calls

[Frits van Bommel]

You seem to have a misunderstanding on the semantics of 'out':

Derek Parnell wrote:
[removed irrelevant lines from below source]
>   void funcOne(in int A, inout int B, out int C)
>   {
>        if (C > 0)  // The value is not readable as 'out' always
>                    // initializes it before the function gets control.
>           C = 3;   // The value is changable and changes
>                    // go back to the caller.
>   }
> 
>   int c = 9;
>   funcOne( a,b,c)
>   //     c is still 9 ('out' prevented called func from seeing
>   //                    its value at call time)

Wrong. c is now 0, because 'out' parameters are initialized at the start 
of the function. When the condition was evaluated c == C == 0 so the 
condition is false and c remains 0, *even after the function returns*.

>   void funcTwo(in char[] A, inout char[] B, out char[] C)
>   {
>        // NB: When passing arrays, the value passed is the reference!
>        if (C.length > 0)  // The value is not readable as 'out' always
>                    // initializes it before the function gets control.
>           C = "c".dup;   // The value is changable and changes
>                    // go back to the caller.
>   }
> 
>   char[] cc = "ccc".dup;
>   funcTwo( a,b,c)
>   //     cc is still "ccc" ('out' prevented called func from seeing
>   //                    its value at call time)

cc should now be null (.length == 0, .ptr == null) for similar reasons 
as in the first example.