return type and templates

[Jonathan M Davis]

On Friday, November 22, 2013 11:50:57 Andrea Fontana wrote:
> I just mean:
> 
> int t = s.value;  // Means  int t = s.value!int;
> 
> If there's a problem with template instantiatio is the same we
> have now.
> Now I have to write:
> 
> int t = s.value!int;
> 
> so if there's a problem with !int, it's just like now.
> 
> It's just a syntactic sugar, no new feature... Am I wrong?

Again, how is the compiler supposed to have any clue that you want to 
instantiate value with int in the case of

int t = s.value;

The left-hand side of the expression has no impact on the type of the right-
hand side, and you have not given the compiler any information as to what 
template argument should be given to value. s.value(3) only works because 
you've given value a function argument from which the corresponding template 
argument can be inforred. With s.value, you've given no indication whatsoever 
as to what value should be instantiated with.

If you want a default template argument, then give it one. e.g.

@property auto value(T = int)() if (is(T == int)) { return _intValue; }

But I don't know how you expect the compiler to have any clue what type value 
should be instantiated with when you haven't given it any template arguments 
and there are no function arguments to infer the template arguments from - 
especially when this what the compiler really sees

template value(T)
    if(is(T == int))
{
    @property auto value() { return _intValue; }
}

and it doesn't even look at the template constraint, let alone the contents of 
the template, until you attempt to instantiate the template. And it's not 
going to be able to try and instantiate the template without having any 
template arguments.

- Jonathan M Davis