out default argument of void

[Timon Gehr]

On 01/04/2012 11:19 PM, Caligo wrote:
> I have a function that looks something like this:
>
> bool fun(double theta, out A a, out B b, out C c){  /* ... */ }
>
> if fun() returns false, then nothing is supposed to be assigned to a,
> b, c.  If it returns true, then values are assigned to a, b, c.  Also,
> there are two ways to call fun():  If I'm interested in the return
> value only, then
>
> 1. fun(theta);
>
> otherwise,
>
> 2. fun(theta, a, b, c);
>
> Obviously, method #1 won't work because there is no such thing as:
>
> bool fun(double theta, out A a = void, out B b = void, out C c =
> void){  /* ... */ }
>
> is there?
>
> So, I have to lose 'out' and use pointers instead:
>
> bool fun(double theta, A* a = null, B* b = null, C* c = null){  /* ... */ }
>
> I don't want to use a variadic function for this either because it's
> not really a variadic function.
>
> 1. Are there any other solutions ?

Yes. Use overloading.

This would work:

bool fun(double theta, out A a, out B b, out C c){  /* ... */ }
bool fun(double theta){
     A a; B b; C c;
     return fun(theta,a,b,c);
}

Or this:

private bool funImpl(double theta, A* a = null, B* b = null, C* c = 
null){  /* ... */ }

bool fun(double theta, out A a, out B b, out C c){return 
funImpl(theta,&a,&b,&c);
bool fun(double theta){return funImpl(theta);}


> 2. Would it make sense to have 'out default argument of void' in D?

void initializer means uninitialized. I don't think that can do what you 
want.