On 4/5/2014 6:26 PM, Michel Fortin wrote:
What if you also have a C++ foo at global scope?
It'll work exactly the same as import does.
module cpptest;
extern (C++) void foo();
extern (C++, namespace = A) void foo();
foo(); // ambiguous
A.foo(); // works
.foo(); // works?
Yes.
cpptest.foo(); // works?
Yes.
Does these two last lines make sense?
Just as much sense as:
module bar;
void foo();
.foo(); // works
bar.foo(); // works
Namespace lookup rules would be exactly the same as for imports and mixin
templates.
