Explicitly saying ref or out when invoking a function

[Lutger]

Ary Borenszweig wrote:

> In C# when you define a function that takes an out or ref parameter,
> when invoking that function you must also specify ref or out. For example:
> 
> void fun(ref uint x, double y);
> 
> uint a = 1;
> double b = 2;
> fun(ref a, b);
> 
> When I first started using C# it really annoyed me that I had to put
> that keyword there just to get my program compiled. "I know what I'm
> doing", I thought. But later, when reading the code, I found it very
> helpful to know that my "a" could be changed when invoking "fun". As
> always, code is read much more times than written, and I think this
> little tips help better understand the code.
> 
> What do you think?

I'm not sure. However, you could implement something like this in descent 
without having to type all that or change D ;) Change the color of the 
parameter or something like that...