On Sunday, 15 September 2013 at 18:31:30 UTC, Marek Janukowicz wrote:
The code to reproduce the problem consists of 3 modules:

mix.d:

module mix;

mixin template A( alias x) {

  string a () {
    return x;
  }
}

====================
aux.d:

module aux;

import mix;

mixin A!("a in aux") X;
string b () { return "b in aux"; }

====================
main.d:

module main;

import aux;
import mix;
import std.stdio;

mixin A!("a in main") X;
string b () { return "b in main"; }

void main () {
  writefln( "a: %s", X.a );  // Line 1
  //writefln( "a: %s", a );  // Line 2
  writefln( "b: %s", b ); // Line 3
}

I run it with: dmd -run main.d aux.d mix.d

Line 1 works. Line 3 works. Line 2 fails with:
main.d(13): Error: main.A!("a in main").a at mix.d(5) conflicts with aux.A!
("a in aux").a at mix.d(5)

If I omit mixin identifier ("X"), there is no way I can make the call to "a"
work without prepending module name.

My question is: why calling a function with the same name (from different
modules) works when:
- it is just a regular function
- it is a mixed-in function with mixin identifier (even though the
identifier is ambiguous)

and it doesn't when it's a mixed-in function with no mixin identifier.

My first impression is that either both line 1 and 2 should work or neither of them should work. It's no surprise to me that line 3 works (and it matches the documentation), so I basically included that just for reference.

Currently this is not a bug.

Looking from the module 'main', the mixin identifier 'X' declared in main.d is *closer* than the 'X' declared in aux.d, because the latter exists beyond the module import boundary. Therefore, the use of 'X' in main.d would prefere the `mixin A!("a in main") X`.

On the other hand, when the name search, all mixed-in symbols are treated as if they are just imported at the mixed-in scope. Therefore, even from main.d, the two mixed-in functions 'a' have same closeness, and the call is ambiguous because they have exactly same signature.

Kenji Hara

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