On 05/14/2013 11:30 AM, Timon Gehr wrote:

On 05/14/2013 10:04 AM, deadalnix wrote:

On Tuesday, 14 May 2013 at 07:28:57 UTC, Timothee Cour wrote:

I think that should be consistent with the deduction mechanism proposed

in the DIP: foo is struct not in scope until template foo(T) is
instantiated.

 No it is not. The DIP states "find all constructors".

it said  'Find all matching class/struct types in scope', isn't that
enough
or am I missing something?

To clarify, I've added 3 out of scope structs named A in the example
section to clarify that they won't be included in the overload set.
see the ones marked '//not in scope'

I think the best here is to specify that the rule is the same than
eponymous funtion and IFTY. Otherwise we'll have 2 different specs with
small difference here and there. And that sucks.

There is no spec for the IFTI case. (i.e. what happens if the eponymous
declaration inside the template is an overload set?)

DMD's strategy is roughly: use the first eponymous declaration that can be found without analysing the template body for IFTI, then use the first eponymous declaration in the analyzed template body to resolve the eponymous declaration after instantiation.

---
import std.stdio;
template fun(T){
        int fun(double arg){ return 1; }
        int fun(T arg){ return 0; }
}
void main(){ writeln(fun(2)); } // error

---
import std.stdio;
template fun(T){
        int fun(T arg){ return 0; }
        int fun(double arg){ return 1; }
}
void main(){ writeln(fun(2)); } // ok

---

This has funny implications, as the compiler may decide to resolve to a different declaration than was used for IFTI later, without doing any kind of overload resolution within the template body.

---
template fun(T){
        int fun(T arg){ return 0; }
        static if(true) int fun(double arg){ return 1; }
}
pragma(msg, fun(2)); // 0
---
template fun(T){
        static if(true) int fun(double arg){ return 1; }
        int fun(T arg){ return 0; }
}
pragma(msg, fun(2)); // 1
---

In the second case, instantiation is performed with the second function, so T is resolved to 'int', but in the end, the 'double' overload is called with an implicit conversion.