On 08/10/2016 12:10 AM, Engine Machine wrote:
I try to use a mixin template and redefine some behaviors but D includes
both and then I get ambiguity. I was sure I read somewhere that when one
uses mixin template it won't include what is already there?
mixin template X
{
void foo() { }
}
struct s
{
mixin template
void foo() { }
}
I was pretty sure I read somewhere that D would not include the foo from
the template since it already exists.
Please post proper code.
This compiles and calls the foo that's not being mixed in:
----
import std.stdio;
mixin template X()
{
void foo() { writeln("mixed in"); }
}
struct s
{
mixin X;
void foo() { writeln("not mixed in"); }
}
void main()
{
s obj;
obj.foo();
}
----
This is in line with the spec, which says: "If the name of a declaration
in a mixin is the same as a declaration in the surrounding scope, the
surrounding declaration overrides the mixin one" [1]. That may be what
you've read.
[1] http://dlang.org/spec/template-mixin.html#mixin_scope
